Hyperbola – JEE Main MathematicsPractice Questions & PYQs

The hyperbola is centered at the origin with vertices at $(\pm 6, 0)$, so its equation takes the form $\frac{x^2}{36} - \frac{y^2}{b^2} = 1$. Substituting the coordinates of the point $P(10, 16)$ allows us to determine $b^2$ as $\frac{100}{36} - \frac{256}{b^2} = 1$, leading to $\frac{256}{b^2} = \frac{64}{36}$, which simplifies to $b^2 = 144$.

The equation of the normal at a point $(x_1, y_1)$ on this hyperbola is expressed by the formula $\frac{a^2 x}{x_1} + \frac{b^2 y}{y_1} = a^2 + b^2$. Substituting the known parameters $a^2 = 36$, $b^2 = 144$, $x_1 = 10$, and $y_1 = 16$ results in $\frac{36x}{10} + \frac{144y}{16} = 36 + 144$, which simplifies to $3.6x + 9y = 180$ and reduces to $2x + 5y = 100$.

The standard equation of a hyperbola is given by
$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$
where the eccentricity relates the semi-axes as $e^2 = 1 + \frac{b^2}{a^2}$. Given an eccentricity of $2$, we have $4 = 1 + \frac{b^2}{a^2}$, which implies $b^2 = 3a^2$, allowing the equation to be expressed as
$\frac{x^2}{a^2} - \frac{y^2}{3a^2} = 1$

Substituting the coordinates $(4, 6)$ into this equation gives $\frac{16}{a^2} - \frac{36}{3a^2} = 1$, or $\frac{16}{a^2} - \frac{12}{a^2} = 1$, which simplifies to $\frac{4}{a^2} = 1$, revealing that $a^2 = 4$. Consequently, the hyperbola is defined by $\frac{x^2}{4} - \frac{y^2}{12} = 1$. The tangent line at any point $(x_1, y_1)$ on this hyperbola is determined by the formula
$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$

Inserting the point $(4, 6)$ and the calculated values for $a^2$ and $b^2$ produces $\frac{4x}{4} - \frac{6y}{12} = 1$. This simplifies to $x - \frac{y}{2} = 1$, which leads to the linear equation $2x - y - 2 = 0$.

The equation of a standard hyperbola is determined by identifying its center, transverse axis, and the constants $a$ and $b$. Since the vertices are at $(-2, 0)$ and $(2, 0)$, the center is the origin $(0, 0)$ and $a=2$. With one focus at $(-3, 0)$, the distance $ae=3$, leading to $e = \frac{3}{2}$. Using the fundamental hyperbola relationship $b^2 = a^2(e^2 - 1)$, we determine $b^2 = 4(\frac{9}{4} - 1)$, which simplifies to $b^2 = 5$.

This results in the standard equation:
$\frac{x^2}{4} - \frac{y^2}{5} = 1$
Substituting each point into this expression confirms which coordinate pair fails the equality. For the point $(6, 5\sqrt{2})$, the equation becomes $\frac{36}{4} - \frac{50}{5}$, which simplifies to $9 - 10$, equaling $-1$ rather than the required $1$.

The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is written as $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Because the length of the transverse axis is 4, we identify that $2a = 4$, which means $a = 2$ and $a^2 = 4$.

By substituting the coordinates of the point (4,2) into the equation $\frac{x^2}{4} - \frac{y^2}{b^2} = 1$, we obtain $4 - \frac{4}{b^2} = 1$. Simplifying this expression leads to $3 = \frac{4}{b^2}$, revealing that $b^2 = \frac{4}{3}$. The eccentricity $e$ is determined by the formula $e = \sqrt{1 + \frac{b^2}{a^2}}$. Plugging in the known values yields $e = \sqrt{1 + \frac{4/3}{4}}$, which simplifies to $\sqrt{1 + \frac{1}{3}}$ or $\sqrt{\frac{4}{3}}$, resulting in a value of $\frac{2}{\sqrt{3}}$.

The eccentricity of a hyperbola is fundamentally linked to the relationship between the semi-axes and the distance to the foci, governed by the following identity:
$b^2 = a^2(e^2 - 1) = (ae)^2 - a^2$
Given that the length of the conjugate axis is 5, the semi-conjugate axis is $b = 2.5$, making $b^2 = \frac{25}{4}$. The total distance between the foci is 13, which corresponds to $2ae = 13$, so the product $ae$ equals $6.5$ or $\frac{13}{2}$.

Substituting these values into the geometric relation, we have $\frac{25}{4} = (\frac{13}{2})^2 - a^2$. Simplifying the right side results in $\frac{25}{4} = \frac{169}{4} - a^2$. Rearranging to solve for $a^2$ yields $a^2 = \frac{169}{4} - \frac{25}{4} = \frac{144}{4} = 36$, so the semi-transverse axis is $a = 6$. With $a = 6$ and $ae = \frac{13}{2}$, the eccentricity $e$ is determined by calculating $\frac{ae}{a} = \frac{13/2}{6}$, which simplifies to $\frac{13}{12}$.

The locus of the center of a circle can be determined by representing the circle with the general form $(x - h)^2 + (y - k)^2 = r^2$, where the center is $(h, k)$ and $r$ is the radius. When this circle intersects the x-axis, the y-coordinate is zero, leading to the equation $(x - h)^2 + k^2 = r^2$, which simplifies to $(x - h)^2 = r^2 - k^2$. The length of the chord on the x-axis is the distance between the two roots of this equation, which is $2\sqrt{r^2 - k^2}$. Since this chord length is given as $4a$, we can equate $2\sqrt{r^2 - k^2} = 4a$, which leads to $\sqrt{r^2 - k^2} = 2a$ or $r^2 - k^2 = 4a^2$, confirming that $r^2 = 4a^2 + k^2$.

Since the circle passes through a point $(0, 2b)$ on the y-axis, we substitute these coordinates into the original circle equation to obtain $h^2 + (2b - k)^2 = r^2$. Expanding this expression yields $h^2 + 4b^2 - 4bk + k^2 = r^2$. Replacing $r^2$ with the expression $4a^2 + k^2$ derived from the chord condition, the equation becomes $h^2 + 4b^2 - 4bk + k^2 = 4a^2 + k^2$. Canceling the $k^2$ terms results in $h^2 - 4bk + 4b^2 - 4a^2 = 0$. By replacing the center coordinates $(h, k)$ with general coordinates $(x, y)$, the locus is described by $x^2 - 4by + 4(b^2 - a^2) = 0$, which is the standard form of a parabola.

A common tangent line to both curves must satisfy their respective geometric conditions, meaning it touches both the parabola and the hyperbola at exactly one point. For the parabola defined by $y^2 = 4x$, any tangent line can be expressed as $y = mx + \frac{1}{m}$ where $m$ represents the slope. Finding the line that also touches the hyperbola $xy = 2$ involves substituting this expression for $y$ into the hyperbola's equation, resulting in the following quadratic equation:
$mx^2 + \frac{x}{m} - 2 = 0$

For this line to be tangent to the hyperbola, the quadratic must have exactly one root, meaning its discriminant must equal zero. Applying the condition $b^2 - 4ac = 0$ yields $(\frac{1}{m})^2 - 4(m)(-2) = 0$, which simplifies to $\frac{1}{m^2} + 8m = 0$. Solving this cubic equation gives $8m^3 = -1$, and therefore the slope must be $m = -\frac{1}{2}$. Inserting this slope value back into the initial tangent formula results in $y = -\frac{1}{2}x - 2$, which rearranges into the final linear form $x + 2y + 4 = 0$.

The hyperbola's geometry links its directrix and focus through the eccentricity $e$, where the directrix is expressed as $x = -\frac{a}{e}$ and the focus as $(-ae, 0)$. By converting the hyperbola equation $16x^2 - 9y^2 = 144$ to the standard form $\frac{x^2}{9} - \frac{y^2}{16} = 1$, we identify the value of the semi-major axis $a = 3$. Given the directrix is $5x + 9 = 0$, we rewrite this as $x = -\frac{9}{5}$ and set it equal to $-\frac{a}{e}$, which is $-\frac{3}{e}$, to solve for the eccentricity $e = \frac{5}{3}$. Multiplying the semi-major axis $a$ by the eccentricity $e$ results in the focal distance $ae = 3 \cdot \frac{5}{3} = 5$, placing the focus at $(-5, 0)$.

A line is tangent to a hyperbola when the intersection of the line and the curve yields a single point of contact, which is determined algebraically by the condition $y = mx \pm \sqrt{a^2m^2 - b^2}$ for a hyperbola in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Parallel lines necessarily share the same slope, allowing us to derive the tangent equation directly from the hyperbola's standard form and the slope of the reference line.

Dividing $4x^2 - 5y^2 = 20$ by 20 yields $\frac{x^2}{5} - \frac{y^2}{4} = 1$, which identifies the parameters $a^2 = 5$ and $b^2 = 4$. Because the requested tangent must be parallel to $x - y = 2$, it inherits the slope $m = 1$. Substituting these values into the tangency condition produces $y = 1x \pm \sqrt{5(1)^2 - 4}$, which simplifies to $y = x \pm 1$. Rearranging this equation into general form results in $x - y \pm 1 = 0$, confirming that $x - y + 1 = 0$ is a valid tangent to the curve.

For a straight line $y = mx + c$ to be normal to a hyperbola defined by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, it must satisfy the specific condition $\frac{a^2}{m^2} - \frac{b^2}{1} = \frac{(a^2 + b^2)^2}{c^2}$. Given the hyperbola equation $\frac{x^2}{24} - \frac{y^2}{18} = 1$, we identify the parameters $a^2 = 24$ and $b^2 = 18$, along with the intercept $c = 7\sqrt{3}$ from the provided normal line. Substituting these values into the condition yields the equation $\frac{24}{m^2} - 18 = \frac{(24 + 18)^2}{(7\sqrt{3})^2}$. Simplifying the right side of the expression, we calculate $\frac{42^2}{49 \times 3}$, which reduces to $\frac{1764}{147} = 12$. Adding $18$ to both sides results in $\frac{24}{m^2} = 30$, which simplifies to $m^2 = \frac{24}{30} = \frac{4}{5}$. Solving for $m$ gives the value $\frac{2}{\sqrt{5}}$.

The hyperbola defined by $\frac{x^2}{\cos^2 \theta} - \frac{y^2}{\sin^2 \theta} = 1$ has semi-axes $a = \cos \theta$ and $b = \sin \theta$. The eccentricity $e$ is calculated using the identity $e = \sqrt{1 + \frac{b^2}{a^2}}$, which simplifies to $\sqrt{1 + \tan^2 \theta} = \sec \theta$. Given the condition that $e > 2$, it follows that $\sec \theta > 2$, which implies $\cos \theta < \frac{1}{2}$. Since $0 < \theta < \frac{\pi}{2}$, this inequality restricts $\theta$ to the interval $(\frac{\pi}{3}, \frac{\pi}{2})$.

The length of the latus rectum is given by the formula $\frac{2b^2}{a}$, which can be expressed in terms of $\theta$ as $\frac{2 \sin^2 \theta}{\cos \theta}$. By substituting $\sin^2 \theta = 1 - \cos^2 \theta$, this expression becomes $2(\frac{1}{\cos \theta} - \cos \theta)$, or $2(\sec \theta - \cos \theta)$. As $\theta$ moves from $\frac{\pi}{3}$ toward $\frac{\pi}{2}$, the value of $\sec \theta$ increases from 2 toward infinity, while $\cos \theta$ decreases from $\frac{1}{2}$ toward 0. Evaluating this expression at the lower boundary $\theta = \frac{\pi}{3}$ yields $2(2 - \frac{1}{2}) = 3$, and because the function is strictly increasing, the length of the latus rectum spans the interval $(3, \infty )$.

For a hyperbola defined by the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the directrix is $x = \frac{a}{e}$, and the relationship between the semi-axes and the eccentricity is $b^2 = a^2(e^2 - 1)$. Given the directrix $5x = 4\sqrt{5}$, we determine $x = \frac{4}{\sqrt{5}}$, which implies $\frac{a}{e} = \frac{4}{\sqrt{5}}$, or $a = \frac{4e}{\sqrt{5}}$. Substituting this into the eccentricity relation gives $b^2 = \frac{16e^2}{5}(e^2 - 1)$.

The hyperbola passes through the point $(4, -2\sqrt{3})$, so substituting these coordinates into the standard equation yields $\frac{16}{a^2} - \frac{12}{b^2} = 1$. Replacing $a^2$ with $\frac{16e^2}{5}$ and $b^2$ with $\frac{16e^2}{5}(e^2 - 1)$ results in the equation $\frac{5}{e^2} - \frac{15}{4e^2(e^2 - 1)} = 1$. Multiplying the entire equation by $4e^2(e^2 - 1)$ yields $20(e^2 - 1) - 15 = 4e^2(e^2 - 1)$, which simplifies to $20e^2 - 35 = 4e^4 - 4e^2$. Rearranging the terms leads to $4e^4 - 24e^2 + 35 = 0$.

The locus of the intersection of a family of lines depending on a parameter $k$ is determined by eliminating $k$ between the equations of the lines. Given the lines $\sqrt{2}x - y + 4\sqrt{2}k = 0$ and $\sqrt{2}kx + ky - 4\sqrt{2} = 0$, the first can be written as $\sqrt{2}x - y = -4\sqrt{2}k$ and the second as $k(\sqrt{2}x + y) = 4\sqrt{2}$, which implies $\sqrt{2}x + y = \frac{4\sqrt{2}}{k}$. Multiplying these two equations eliminates $k$: $(\sqrt{2}x - y)(\sqrt{2}x + y) = (-4\sqrt{2}k) \left( \frac{4\sqrt{2}}{k} \right)$, which simplifies to $2x^2 - y^2 = -32$. Rearranging this result into the standard hyperbola form yields $y^2 - 2x^2 = 32$, or equivalently, $\frac{y^2}{32} - \frac{x^2}{16} = 1$. This equation represents a hyperbola with its transverse axis along the $y$-axis, where $a^2 = 32$, so $a = \sqrt{32} = 4\sqrt{2}$. The length of the transverse axis is $2a = 8\sqrt{2}$.

The equation of the hyperbola is $4y^2 - x^2 = 1$. The equation of the tangent at any point $(x_1, y_1)$ on this hyperbola is $4y y_1 - x x_1 = 1$. The intersection points of this tangent with the coordinate axes are determined by setting $y=0$ to find A$(-1/x_1, 0)$ and setting $x=0$ to find B$(0, 1/(4y_1))$. The coordinates of the midpoint $(h, k)$ of AB are given by $h = -1/(2x_1)$ and $k = 1/(8y_1)$, which imply $x_1 = -1/(2h)$ and $y_1 = 1/(8k)$. Substituting these coordinates into the hyperbola equation $4y_1^2 - x_1^2 = 1$ yields $4(1/(8k))^2 - (-1/(2h))^2 = 1$. Simplifying this gives $\frac{1}{16k^2} - \frac{1}{4h^2} = 1$, and multiplying the entire equation by $16h^2k^2$ leads to $h^2 - 4k^2 = 16h^2k^2$. Replacing $(h, k)$ with $(x, y)$, the locus is $x^2 - 4y^2 - 16x^2y^2 = 0$.

The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec \theta, b \tan \theta)$ is given by $\frac{ax}{\sec \theta} + \frac{by}{\tan \theta} = a^2 + b^2$. For the given hyperbola $4x^2 - 9y^2 = 36$, which simplifies to $\frac{x^2}{9} - \frac{y^2}{4} = 1$, we have $a=3$ and $b=2$, making the normal equation $\frac{3x}{\sec \theta} + \frac{2y}{\tan \theta} = 13$. The $x$-intercept $A$ is obtained by setting $y=0$, resulting in $x_A = \frac{13 \sec \theta}{3}$, and the $y$-intercept $B$ is obtained by setting $x=0$, resulting in $y_B = \frac{13 \tan \theta}{2}$. In the parallelogram $OABP$ where $O$ is the origin, the coordinates of vertex $P(x, y)$ correspond to $(x_A, y_B)$, thus $x = \frac{13 \sec \theta}{3}$ and $y = \frac{13 \tan \theta}{2}$. Rearranging these to isolate the trigonometric terms yields $\sec \theta = \frac{3x}{13}$ and $\tan \theta = \frac{2y}{13}$. Applying the fundamental identity $\sec^2 \theta - \tan^2 \theta = 1$, we substitute these expressions to get $(\frac{3x}{13})^2 - (\frac{2y}{13})^2 = 1$, which simplifies to $9x^2 - 4y^2 = 169$.

The line segment connecting the points of tangency $P$ and $Q$ is defined as the chord of contact for the hyperbola $4x^2 - y^2 = 36$ originating from the external point $T(0, 3)$. Using the principle $T=0$ for the hyperbola, the equation of this chord is $4x(0) - y(3) - 36 = 0$, which reduces to the horizontal line $y = -12$. Substituting $y = -12$ back into the original hyperbola equation yields $4x^2 - (-12)^2 = 36$, resulting in $4x^2 = 180$, so $x = \pm 3\sqrt{5}$. With vertices at $T(0, 3)$, $P(3\sqrt{5}, -12)$, and $Q(-3\sqrt{5}, -12)$, the triangle features a horizontal base $PQ$ of length $6\sqrt{5}$ and a vertical altitude measured from $y = -12$ to $y = 3$, giving a height of 15. The area is calculated as $\frac{1}{2} \times 6\sqrt{5} \times 15$, which equals $45\sqrt{5}$ square units.

Finding the locus of intersection points for two lines that depend on a variable parameter requires eliminating that parameter to relate the coordinates $x$ and $y$ directly. Given the equations $tx-2y-3t=0$ and $x-2ty+3=0$, we rearrange the first to get $t(x-3)=2y$, which implies $t=\frac{2y}{x-3}$, and rewrite the second as $x+3=2ty$. Substituting the expression for $t$ into the second equation yields $x+3=2\left(\frac{2y}{x-3}\right)y$. Multiplying both sides by $x-3$ leads to the identity $(x+3)(x-3)=4y^2$, which simplifies to $x^2-9=4y^2$. Rearranging this relationship into the standard form of a hyperbola, $\frac{x^2}{9} - \frac{y^2}{9/4} = 1$, reveals the properties of the curve where $a^2=9$ and $b^2=\frac{9}{4}$. Since the length of the conjugate axis is $2b$, calculating $2\sqrt{\frac{9}{4}}$ results in $2 \times \frac{3}{2}$, which confirms the length of the conjugate axis is 3.

The standard equation for a hyperbola with foci at $(\pm 2, 0)$ is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. Given the foci, we have $ae = 2$, which leads to $a^2e^2 = 4$. Applying the hyperbola property $b^2 = a^2(e^2 - 1)$, we substitute $a^2e^2 = 4$ to get $b^2 = 4 - a^2$, or $a^2 + b^2 = 4$.

Since the hyperbola passes through the point $(\sqrt{2}, \sqrt{3})$, the coordinates must satisfy $\frac{2}{a^2} - \frac{3}{b^2} = 1$. Substituting $b^2 = 4 - a^2$ into this equation gives $\frac{2}{a^2} - \frac{3}{4 - a^2} = 1$. Simplifying this expression results in the quadratic $a^4 - 9a^2 + 8 = 0$, which has roots $a^2 = 1$ and $a^2 = 8$. Because $b^2 = 4 - a^2$ must be positive, we reject $a^2 = 8$ and conclude $a^2 = 1$ and $b^2 = 3$.

The equation of the hyperbola is therefore $x^2 - \frac{y^2}{3} = 1$. The tangent line at any point $(x_0, y_0)$ is given by the formula $\frac{x x_0}{a^2} - \frac{y y_0}{b^2} = 1$. Substituting the point $(\sqrt{2}, \sqrt{3})$ and our values for $a^2$ and $b^2$, the tangent equation becomes $\sqrt{2}x - \frac{\sqrt{3}y}{3} = 1$, which simplifies to $\sqrt{2}x - \frac{y}{\sqrt{3}} = 1$. Testing the provided coordinates, we find that the point $(2\sqrt{2}, 3\sqrt{3})$ lies on this tangent line because $\sqrt{2}(2\sqrt{2}) - \frac{3\sqrt{3}}{\sqrt{3}} = 4 - 3 = 1$.

The equation $9e^2-18e+5=0$ yields $e=\frac{5}{3}$ because hyperbola eccentricity must exceed one. We identify the distance from the center to the focus as $ae=5$ and from the center to the directrix as $\frac{a}{e}=\frac{9}{5}$, meaning their difference is $a(e-\frac{1}{e}) = 5-\frac{9}{5} = \frac{16}{5}$. Substituting $e=\frac{5}{3}$ into $a(\frac{5}{3}-\frac{3}{5}) = \frac{16}{5}$ simplifies to $a=3$. Using the identity $b^2 = a^2(e^2-1)$, we calculate $b^2 = 9(\frac{25}{9}-1) = 16$, which makes $a^2-b^2$ equal to $9-16$, or $-7$.

The length of the latus rectum for a standard hyperbola is given by the formula $\frac{2b^2}{a} = 8$, which simplifies to $b^2 = 4a$. The conjugate axis has a length of $2b$, and the distance between the foci is $2ae$; equating the conjugate axis to half this distance yields $2b = ae$. Squaring this expression gives $4b^2 = a^2e^2$, and substituting $b^2 = 4a$ transforms this into $16 = ae^2$. By applying the standard hyperbola relationship $b^2 = a^2(e^2 - 1)$ and substituting $b^2 = 4a$, we obtain $a = \frac{4}{e^2 - 1}$. Substituting this value of $a$ into $16 = ae^2$ results in $16 = \frac{4e^2}{e^2 - 1}$, which simplifies to $4(e^2 - 1) = e^2$. This equation reduces to $3e^2 = 4$, producing an eccentricity of $e = \frac{2}{\sqrt{3}}$.