Hyperbola – JEE Main MathematicsPractice Questions & PYQs

To determine the properties of the hyperbola, we use the condition of tangency for a line $y = mx + c$ on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, which is $c^2 = a^2m^2 - b^2$. The given tangent line is $y - \sqrt{3}x + \sqrt{3} = 0$, which can be rewritten as $y = \sqrt{3}x - \sqrt{3}$, identifying the slope $m = \sqrt{3}$ and intercept $c = -\sqrt{3}$. Substituting these into the tangency condition yields $3 = 3a^2 - b^2$. We also know the length of the latus rectum is $\frac{2b^2}{a} = 9$, or $b^2 = \frac{9a}{2}$. By substituting this expression for $b^2$ into our tangency equation, we obtain $3 = 3a^2 - \frac{9a}{2}$, which simplifies to the quadratic $2a^2 - 3a - 2 = 0$. Factoring this equation as $(2a+1)(a-2) = 0$ reveals that $a = 2$, and subsequently, $b^2 = 9$.

With the hyperbola defined as $\frac{x^2}{4} - \frac{y^2}{9} = 1$, its eccentricity is calculated using $e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{9}{4} = \frac{13}{4}$. The point of contact $(x_0, y_0)$ is the point where the line $y = \sqrt{3}x - \sqrt{3}$ touches the hyperbola. Comparing this line to the standard tangent equation $\frac{xx_0}{a^2} - \frac{yy_0}{b^2} = 1$, which is $\frac{x x_0}{4} - \frac{y y_0}{9} = 1$, or $x - \frac{y}{\sqrt{3}} = 1$, we identify $x_0 = 4$ and $y_0 = 3\sqrt{3}$.

The product of the focal distances $m$ for any point $(x_0, y_0)$ on a hyperbola is given by the expression $|e^2x_0^2 - a^2|$. Substituting our calculated values of $x_0 = 4$, $e^2 = \frac{13}{4}$, and $a^2 = 4$ into this formula gives $m = \frac{13}{4}(4^2) - 4 = 13(4) - 4 = 52 - 4 = 48$. Finally, computing the required value $4e^2 + m$, we find $4(\frac{13}{4}) + 48 = 13 + 48 = 61$.

For the hyperbola $\frac{x^2}{3} - \frac{y^2}{5} = 1$, the distance $c$ from the center to the focus is determined by $\sqrt{a^2 + b^2} = \sqrt{3 + 5} = \sqrt{8}$. Given that the focus $S$ lies on the positive $x$-axis, its coordinates are $(\sqrt{8}, 0)$. Since $A(\sqrt{6}, \sqrt{5})$ is the center of the circle and $SAB$ forms a diameter, the center $A$ is the midpoint of the segment $SB$. Applying the midpoint relation $A = \frac{S+B}{2}$ identifies the coordinates of point $B$ as $(2\sqrt{6} - \sqrt{8}, 2\sqrt{5})$.

In triangle $OSB$, the base $OS$ lies along the $x$-axis and has a length of $\sqrt{8}$. The height of the triangle is the vertical distance from the base to point $B$, which is simply the $y$-coordinate of $B$, or $2\sqrt{5}$. The area is calculated as $\frac{1}{2} \cdot \text{base} \cdot \text{height} = \frac{1}{2} \cdot \sqrt{8} \cdot 2\sqrt{5} = \sqrt{40}$. Squaring this value gives the final result of 40.

The hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$ possesses foci at $(\pm 5, 0)$ derived from $c = \sqrt{16+9}$, resulting in an eccentricity of $e_1 = \frac{5}{4}$. Given that $e_1 e_2 = 1$, the ellipse must have an eccentricity of $e_2 = \frac{4}{5}$. Since the ellipse passes through the hyperbola's foci $(\pm 5, 0)$ and satisfies $a > b$, its semi-major axis is $a = 5$, and solving $e_2^2 = 1 - \frac{b^2}{a^2}$ yields $\frac{16}{25} = 1 - \frac{b^2}{25}$, which gives $b^2 = 9$ and the ellipse equation $\frac{x^2}{25} + \frac{y^2}{9} = 1$. Substituting $y = 2$ into this equation results in $\frac{x^2}{25} = 1 - \frac{4}{9} = \frac{5}{9}$, meaning $x = \pm \frac{5 \sqrt{5}}{3}$. Calculating the difference between these two $x$-coordinates gives the total chord length of $\frac{10 \sqrt{5}}{3}$.

The eccentricity of a conic section is determined by the relationship between its semi-axes, using the formulas $e = \sqrt{1 + \frac{b^2}{a^2}}$ for a hyperbola and $e = \sqrt{1 - \frac{b^2}{a^2}}$ for an ellipse where the major axis is identified by the larger denominator. By rewriting the hyperbola equation $x^2 - y^2 \csc^2 \theta = 5$ into the standard form $\frac{x^2}{5} - \frac{y^2}{5 \sin^2 \theta} = 1$, we find its eccentricity is $e_h = \sqrt{1 + \sin^2 \theta}$. Similarly, transforming the ellipse equation $x^2 \csc^2 \theta + y^2 = 5$ gives $\frac{x^2}{5 \sin^2 \theta} + \frac{y^2}{5} = 1$, which, because $\csc^2 \theta > 1$ for the given range of $\theta$, results in an eccentricity of $e_c = \sqrt{1 - \sin^2 \theta}$.

Given the relationship $e_h = \sqrt{7} e_c$, squaring both sides of the equation allows us to equate the expressions directly:
$1 + \sin^2 \theta = 7(1 - \sin^2 \theta)$
Expanding the right side yields $1 + \sin^2 \theta = 7 - 7 \sin^2 \theta$, which simplifies to $8 \sin^2 \theta = 6$, or $\sin^2 \theta = \frac{3}{4}$. Taking the square root for the first quadrant gives $\sin \theta = \frac{\sqrt{3}}{2}$, confirming the value of $\theta$ is $\frac{\pi}{3}$.

A hyperbola centered at the origin with its foci on the $x$-axis is defined by the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, where the distance from the center to each vertex is $a$. Because circle $C_1$ is centered at the origin and touches the hyperbola at its vertices, its radius must be equal to $a$. Given that the area of this circle is $36 \pi$, we set $\pi a^2 = 36 \pi$, which identifies the semi-major axis as $a = 6$.

Circle $C_2$ is centered at one of the foci, located at $(ae, 0)$, and touches the hyperbola at the vertex $(a, 0)$, meaning its radius is the distance $|ae - a| = a(e-1)$. Since the area of $C_2$ is $4 \pi$, its radius must be $2$, leading to the equation $6(e-1) = 2$, which solves to an eccentricity of $e = \frac{4}{3}$. We then use the hyperbola property $b^2 = a^2(e^2 - 1)$ to find $b^2 = 36 \left(\left(\frac{4}{3}\right)^2 - 1\right) = 36 \left(\frac{16}{9} - 1\right) = 28$. The length of the latus rectum is defined as $\frac{2b^2}{a}$, which results in $\frac{2 \times 28}{6} = \frac{28}{3}$.

The relationship between the eccentricities of the ellipse $E$ and the hyperbola $H$ is foundational, where the hyperbola eccentricity $e_H$ is defined as the reciprocal of the ellipse eccentricity $e_E$. For the ellipse given by $\frac{(x-1)^2}{100}+\frac{(y-1)^2}{75}=1$, the values $a^2=100$ and $b^2=75$ yield an eccentricity $e_E = \sqrt{1-\frac{75}{100}} = \frac{1}{2}$, which directly implies $e_H = 2$. With the ellipse center at $(1,1)$, the focal distance is $c = \sqrt{100-75} = 5$, placing the foci at $(6,1)$ and $(-4,1)$.

These coordinates serve as the foci for the hyperbola $H$, meaning its focal distance is $c_H = 5$. Using the relation $c_H = a_H e_H$, we calculate $5 = a_H(2)$, which gives $a_H = 2.5$, making the length of the transverse axis $\alpha = 2a_H = 5$. Applying the hyperbola identity $c_H^2 = a_H^2 + b_H^2$, we find $b_H^2 = 25 - (2.5)^2 = 18.75$, resulting in a conjugate axis length of $\beta = 2\sqrt{18.75} = 5\sqrt{3}$.

Substituting these values into the required expression, we calculate $3\alpha^2 + 2\beta^2 = 3(5^2) + 2(5\sqrt{3})^2$. This simplifies to $3(25) + 2(75) = 75 + 150 = 225$.

Two conics intersecting at right angles are confocal, meaning they share the same foci. Starting with the hyperbola $2x^2 - 2y^2 = 1$, we rewrite it as $x^2 - y^2 = \frac{1}{2}$, revealing a rectangular hyperbola where $a'^2 = \frac{1}{2}$ and $b'^2 = \frac{1}{2}$. The eccentricity of this hyperbola is calculated as $e' = \sqrt{1 + \frac{b'^2}{a'^2}} = \sqrt{2}$. Consequently, the ellipse has an eccentricity $e = \frac{1}{\sqrt{2}}$. Using the ellipse eccentricity formula $e^2 = 1 - \frac{b^2}{a^2}$, we establish the relationship between its semi-axes:
$1 - \frac{b^2}{a^2} = \frac{1}{2} \implies a^2 = 2b^2$
Because the curves are confocal, the focus of the ellipse $ae$ must equal the focus of the hyperbola $\sqrt{a'^2 + b'^2}$. Since the hyperbola's focus distance is $\sqrt{\frac{1}{2} + \frac{1}{2}} = 1$, we set $ae = 1$. Substituting the ellipse's eccentricity $e = \frac{1}{\sqrt{2}}$ gives $a \cdot \frac{1}{\sqrt{2}} = 1$, which implies $a = \sqrt{2}$. From the earlier relationship $a^2 = 2b^2$, we find that $2 = 2b^2$, so $b^2 = 1$. The length of the latus-rectum is determined by the formula:
$\text{LR} = \frac{2b^2}{a} = \frac{2}{\sqrt{2}} = \sqrt{2}$
Squaring this length yields 2.

The equation of any tangent with slope $m$ to the hyperbola $\frac{y^2}{25} - \frac{x^2}{16} = 1$ is $y = mx \pm \sqrt{25 - 16m^2}$. Substituting the point $P(4, 1)$ into this equation gives $1 = 4m \pm \sqrt{25 - 16m^2}$, which leads to $(1-4m)^2 = 25 - 16m^2$. Expanding and simplifying this expression results in $1 - 8m + 16m^2 = 25 - 16m^2$, which further reduces to $32m^2 - 8m - 24 = 0$, or $4m^2 - m - 3 = 0$. Factoring this quadratic equation yields $(4m+3)(m-1) = 0$, meaning the tangents from $P$ have slopes $m_1 = 1$ and $m_2 = -3/4$.

The tangents from point $Q$ must have slopes $|m_1| = 1$ and $|m_2| = 3/4$. Using the tangency condition again for these slopes, the relevant equations are $y = x \pm \sqrt{25 - 16(1)^2} = x \pm 3$ and $y = \frac{3}{4}x \pm \sqrt{25 - 16(3/4)^2} = \frac{3}{4}x \pm 4$. To obtain positive $x$-intercepts $\alpha$ and $\beta$, we set $y = 0$ for both equations. The line $y = x - 3$ gives an $x$-intercept $\beta = 3$, and the line $y = \frac{3}{4}x - 4$ gives an $x$-intercept $\alpha = 16/3$. These two lines intersect at point $Q$, which we find by solving $x - 3 = \frac{3}{4}x - 4$, resulting in $\frac{1}{4}x = -1$, or $x = -4$, and consequently $y = -7$. With $P(4, 1)$ and $Q(-4, -7)$, the squared distance is $PQ^2 = (4 - (-4))^2 + (1 - (-7))^2 = 8^2 + 8^2 = 128$. Finally, the value of $\frac{(PQ)^2}{\alpha\beta}$ is $\frac{128}{(16/3) \times 3} = \frac{128}{16} = 8$.

Completing the square for the hyperbola equation $16 x^2 - y^2 + 64 x + 4 y + 44 = 0$ begins by grouping terms as $16(x^2+4x) - (y^2-4y) + 44 = 0$. Adjusting these groups leads to $16(x+2)^2 - 64 - (y-2)^2 + 4 + 44 = 0$, which simplifies to $16(x+2)^2 - (y-2)^2 = 16$. Dividing the equation by 16 yields $\frac{(x+2)^2}{1} - \frac{(y-2)^2}{16} = 1$, which defines the transverse axis $T$ as the horizontal line $y=2$ and the conjugate axis $C$ as the vertical line $x=-2$.

The region of interest is bounded above by $y=2$, to the right of $x=-2$, and above the parabola $y=x^2-4$. The boundaries for the definite integral are determined by the intersection of the parabola $y=x^2-4$ and the transverse axis $y=2$, which occurs at $x^2-4=2$, giving $x=\pm\sqrt{6}$. Since the region lies to the right of the conjugate axis $x=-2$, the integration interval is $[-2, \sqrt{6}]$.

The area is calculated by evaluating the integral $\int_{-2}^{\sqrt{6}} (2 - (x^2 - 4)) \, dx$, which simplifies to $\int_{-2}^{\sqrt{6}} (6 - x^2) \, dx$. Computing the antiderivative yields $\left[ 6x - \frac{x^3}{3} \right]_{-2}^{\sqrt{6}}$. Substituting the bounds results in $\left( 6\sqrt{6} - \frac{(\sqrt{6})^3}{3} \right) - \left( 6(-2) - \frac{(-2)^3}{3} \right)$, which simplifies to $(6\sqrt{6} - 2\sqrt{6}) - (-12 + \frac{8}{3})$. Finalizing the arithmetic, the expression becomes $4\sqrt{6} - (- \frac{28}{3})$, resulting in a total area of $4\sqrt{6} + \frac{28}{3}$.

The distance between the foci of a hyperbola is $2ae$, where $a$ represents the semi-transverse axis and $e$ is the eccentricity. Calculating the distance between $(1+\sqrt{2}, 0)$ and $(1-\sqrt{2}, 0)$ gives $2\sqrt{2}$, and setting this equal to $2ae$ with $e=\sqrt{2}$ yields
$2a\sqrt{2} = 2\sqrt{2}$
which simplifies to $a = 1$. The value of $b^2$ follows from the hyperbola relation $b^2 = a^2(e^2 - 1)$, resulting in $b^2 = 1^2(2 - 1) = 1$. Substituting these values into the formula for the length of the latus rectum produces
$\frac{2b^2}{a} = \frac{2(1)}{1} = 2$

The point on a hyperbola that is nearest to a given line is the point where the tangent to the hyperbola is parallel to that line. To find this point on the hyperbola $3x^2 - 4y^2 = 36$, we first determine the slope of the given line $3x + 2y = 1$ by rewriting it as $y = -\frac{3}{2}x + \frac{1}{2}$, which indicates the slope is $-\frac{3}{2}$. We then find the slope of the tangent to the hyperbola at any point $(x_0, y_0)$ by implicit differentiation, yielding $6x_0 - 8y_0 \frac{dy}{dx} = 0$, or $\frac{dy}{dx} = \frac{3x_0}{4y_0}$.

Setting the tangent slope equal to the line slope, $\frac{3x_0}{4y_0} = -\frac{3}{2}$, simplifies to the condition $x_0 = -2y_0$. Substituting this expression for $x_0$ back into the original hyperbola equation $3x_0^2 - 4y_0^2 = 36$ gives $3(-2y_0)^2 - 4y_0^2 = 36$, which reduces to $12y_0^2 - 4y_0^2 = 36$ or $8y_0^2 = 36$. This results in $y_0^2 = \frac{9}{2}$, providing two candidate points $(\frac{6}{\sqrt{2}}, -\frac{3}{\sqrt{2}})$ and $(-\frac{6}{\sqrt{2}}, \frac{3}{\sqrt{2}})$. By evaluating the perpendicular distance from these points to the line $3x + 2y - 1 = 0$, we find that the point $(\frac{6}{\sqrt{2}}, -\frac{3}{\sqrt{2}})$ is the one closer to the line. With $x_0 = \frac{6}{\sqrt{2}}$ and $y_0 = -\frac{3}{\sqrt{2}}$, the expression $\sqrt{2}(y_0 - x_0)$ becomes $\sqrt{2}(-\frac{3}{\sqrt{2}} - \frac{6}{\sqrt{2}})$, which simplifies to $\sqrt{2}(-\frac{9}{\sqrt{2}})$ and equals $-9$.

The rectangle defined by the lines $x=0, x=2, y=0$, and $y=5$ has a total area of $10$, so a line segment $AB$ dividing this area in a $4:1$ ratio must cut off a right-angled triangle $\Delta OAB$ with an area of $2$. Since the area of $\Delta OAB$ is given by $\frac{1}{2}\alpha\beta$, equating this to $2$ yields the relationship $\alpha\beta = 4$. Representing the midpoint $M(h, k)$ of segment $AB$ as $\left(\frac{\alpha}{2}, \frac{\beta}{2}\right)$, we find that $\alpha = 2h$ and $\beta = 2k$, which implies that $(2h)(2k) = 4$ or $hk = 1$. This final equation describes the path of a hyperbola.

The eccentricity $e$ of a hyperbola defined by the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is calculated using the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$. Given the specific hyperbola equation $\frac{x^2}{1+n} - \frac{y^2}{3+n} = 1$, we can substitute $a^2 = 1+n$ and $b^2 = 3+n$ into the expression to find:
$e = \sqrt{1 + \frac{3+n}{1+n}} = \sqrt{\frac{2n+4}{n+1}}$
For the eccentricity to be a rational number, the expression under the square root must evaluate to the square of a rational number. By testing even values for $n$, we determine that the smallest integer that satisfies this condition is $n=48$, resulting in $e = \sqrt{\frac{100}{49}} = \frac{10}{7}$.

With $n=48$, the parameters of the hyperbola are $a = \sqrt{49} = 7$ and $b^2 = 51$. The length of the latus rectum $l$ is calculated using the formula $l = \frac{2b^2}{a}$, which yields:
$l = \frac{2 \times 51}{7} = \frac{102}{7}$
Calculating the final value required, we find that $21l = 21 \times \frac{102}{7} = 3 \times 102 = 306$.

The geometry of a hyperbola is uniquely determined by its foci at $(\pm 2, 0)$ and eccentricity $e = \frac{3}{2}$. Since the foci are positioned at $(\pm Ae, 0)$, we identify the semi-transverse axis $A$ as $2 / e = \frac{4}{3}$. Using the relation $B^2 = A^2(e^2 - 1)$, the square of the semi-conjugate axis is $B^2 = \left(\frac{16}{9}\right)\left(\frac{9}{4} - 1\right) = \frac{20}{9}$, which defines the hyperbola equation as $\frac{x^2}{16/9} - \frac{y^2}{20/9} = 1$.

Any tangent to this hyperbola with slope $m$ follows the form $y = mx \pm \sqrt{A^2m^2 - B^2}$. Given that the tangent is perpendicular to the line $2x + 3y = 6$, which has a slope of $-\frac{2}{3}$, the tangent slope must be $m = \frac{3}{2}$. Substituting these values into the tangent equation yields $y = \frac{3}{2}x \pm \sqrt{\left(\frac{16}{9}\right)\left(\frac{9}{4}\right) - \frac{20}{9}}$, which simplifies to $y = \frac{3}{2}x \pm \frac{4}{3}$. Setting $y = 0$ identifies the x-intercept $a$ as $\pm \frac{8}{9}$, and setting $x = 0$ identifies the y-intercept $b$ as $\pm \frac{4}{3}$. Substituting these values into the expression $|6a| + |5b|$ produces $|6(\frac{8}{9})| + |5(\frac{4}{3})| = \frac{16}{3} + \frac{20}{3} = \frac{36}{3}$, resulting in a final value of $12$.

The fundamental condition for a line with slope $m$ to be tangent to the hyperbola $\frac{x^2}{16} - \frac{y^2}{4} = 1$ is given by $y = mx \pm \sqrt{16m^2 - 4}$. Since line $L_2$ passes through the origin and is perpendicular to $L_1$, its slope must be the negative reciprocal of the slope of $L_1$, which is $-\frac{1}{m}$. The equation of this perpendicular line is thus $y = -\frac{x}{m}$, which allows us to define the slope of the tangent in terms of the intersection coordinates $(x, y)$ as $m = -\frac{x}{y}$.

By substituting $m = -\frac{x}{y}$ back into the tangent equation, we establish the relationship between the coordinates of the intersection point. Rewriting the tangent equation as $y - mx = \pm \sqrt{16m^2 - 4}$ and squaring both sides yields $(y - mx)^2 = 16m^2 - 4$. Inserting $m = -\frac{x}{y}$ into this expression results in $(y - (-\frac{x}{y})x)^2 = 16(-\frac{x}{y})^2 - 4$, which simplifies to $(y + \frac{x^2}{y})^2 = \frac{16x^2}{y^2} - 4$. Clearing the denominator by multiplying through by $y^2$ gives $\frac{(y^2 + x^2)^2}{y^2} = \frac{16x^2 - 4y^2}{y^2}$, which leads directly to the locus $(x^2 + y^2)^2 = 16x^2 - 4y^2$. Comparing this result to the given form $(x^2 + y^2)^2 = \alpha x^2 + \beta y^2$ identifies $\alpha = 16$ and $\beta = -4$, resulting in a sum of $16 - 4 = 12$.

The center of the circle given by $x^2 + y^2 - 2x + 2fy + 1 = 0$ is found by completing the square to rewrite the equation as $(x-1)^2 + (y+f)^2 = f^2$, which identifies the center at $(1, -f)$. Since the circle's diameters must intersect at its center, substituting these coordinates into the equations $2px - y = 1$ and $2x + py = 4p$ results in the simultaneous equations $2p + f = 1$ and $2 - pf = 4p$. Substituting $f = 1 - 2p$ from the first equation into the second yields $2 - p(1 - 2p) = 4p$, which simplifies to the quadratic $2p^2 - 5p + 2 = 0$. Solving this factors into $(2p - 1)(p - 2) = 0$, giving solutions for $p$ as $1/2$ (with $f = 0$) or $2$ (with $f = -3$), corresponding to centers at $(1, 0)$ and $(1, 3)$, respectively.

The hyperbola $3x^2 - y^2 = 3$ can be expressed as $x^2 - \frac{y^2}{3} = 1$, where the equation of any tangent with slope $m$ is given by $y = mx \pm \sqrt{m^2 - 3}$. Passing this tangent through the center $(1, 3)$ results in $3 = m(1) \pm \sqrt{m^2 - 3}$, which rearranges to $3 - m = \pm \sqrt{m^2 - 3}$. Squaring both sides of this equation produces $9 - 6m + m^2 = m^2 - 3$, and simplifying the remaining terms leads to $6m = 12$, confirming that the slope $m$ is equal to 2.

The hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ has an eccentricity $e$ satisfying $e^2 = 1 + \frac{b^2}{a^2}$ and a latus rectum length $l = \frac{2b^2}{a}$. Its conjugate hyperbola, given by $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$, has an eccentricity $e'$ satisfying $(e')^2 = 1 + \frac{a^2}{b^2}$ and a latus rectum length $l' = \frac{2a^2}{b}$. Applying the given conditions $e^2 = \frac{11}{14}l$ and $(e')^2 = \frac{11}{8}l'$ leads to the following two equations:
$a^2 + b^2 = \frac{11b^2 a}{7}$
$a^2 + b^2 = \frac{11a^2 b}{4}$

Equating these two expressions results in $\frac{11b^2 a}{7} = \frac{11a^2 b}{4}$, which simplifies to $4b = 7a$. Substituting $a = \frac{4b}{7}$ into the first equation gives $\left(\frac{4b}{7}\right)^2 + b^2 = \frac{11b^2}{7} \cdot \left(\frac{4b}{7}\right)$, which simplifies to $\frac{16b^2}{49} + b^2 = \frac{44b^3}{49}$. Combining terms yields $\frac{65b^2}{49} = \frac{44b^3}{49}$, which identifies $44b = 65$. Since $7a = 4b$, it follows that $77a = 44b = 65$. The sum $77a + 44b$ is therefore $65 + 65 = 130$.