Question:
The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 $ If the eccentricity of the hyperbola is 2, then the equation of the tangent to this hyperbola passing through the point (4, 6) is :
The foci of a hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 $ If the eccentricity of the hyperbola is 2, then the equation of the tangent to this hyperbola passing through the point (4, 6) is :
Updated On: Nov 21, 2025
- 3x - 2y = 0
- 2x - 3y + 10 = 0
- x - 2y + 8 = 0
- 2x - y - 2 = 0
Hide Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Answer (c) x - 2y + 8 = 0
Was this answer helpful?
0
0
Top Questions on Hyperbola
- Let the foci of a hyperbola be \( (1, 14) \) and \( (1, -12) \). If it passes through the point \( (1, 6) \), then the length of its latus-rectum is:
- Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be $ 2a $ and $ 2b $, respectively, and one focus and the corresponding directrix of this hyperbola be $ (-5, 0) $ and $ 5x + 9 = 0 $, respectively. If the product of the focal distances of a point $ (\alpha, 2\sqrt{5}) $ on the hyperbola is $ p $, then $ 4p $ is equal to:
- If the normal drawn to the hyperbola \( xy = 16 \) at (8, 2) meets the hyperbola again at a point \((\alpha, \beta)\), then \( |\beta| + \frac{1}{|\alpha|} = \)
- If \( 3\sqrt{2}x - 4y = 12 \) is a tangent to the hyperbola \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) and \(\frac{5}{4}\) is its eccentricity, then \( a^2 - b^2 = \)
- The length of the latus rectum and directrices of a hyperbola with eccentricity $e$ are 9 and $x = \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y - \sqrt{3}x + \sqrt{3} = 0$ touch this hyperbola at $(x_0, y_0)$. If $m$ is the product of the focal distances of the point $(x_0, y_0)$, then $4e^2 + m$ is equal to ________.
View More Questions
Questions Asked in JEE Main exam
In the given circuit diagram, $I_L = 5$ mA and the Zener voltage is $V_Z = 5$ V. If $i_z = 4 i_L$, find the value of $R_s$ (in $\Omega$).
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
- For the given statements below, mark the correct option. Statement-I: Work done by a conservative force \( f \), from \( r_1 \) to \( r_2 \) is given by: \[ W = \int_{r_1}^{r_2} f \, dr. \] Statement-II: Work done by conservative force is path dependent.
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
- A metallic conductor of length 2m and cross-sectional area \( 0.2 \, \text{mm}^2 \) carries a steady current of 1.2 A when a potential difference of 2 V is applied across it. (\( e = 1.6 \times 10^{-19} \), charge density = \( 7.5 \times 10^{28} \, \text{m}^{-3} \)) Then the mobility of the charge carrier is \( x \times 10^{-4} \) SI units. Find \( x \).
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
- For given logic gate circuit choose correct truth table.
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
Three silicon diodes connected parallel to each other as shown. Forward voltage of diode is 0.7 V. Find current through diode A :
- JEE Main - 2026
- Semiconductor electronics: materials, devices and simple circuits
View More Questions
Concepts Used:
Hyperbola
Hyperbola is the locus of all the points in a plane such that the difference in their distances from two fixed points in the plane is constant.
Hyperbola is made up of two similar curves that resemble a parabola. Hyperbola has two fixed points which can be shown in the picture, are known as foci or focus. When we join the foci or focus using a line segment then its midpoint gives us centre. Hence, this line segment is known as the transverse axis.
