Question:
If \( 4x - 3y - 5 = 0 \) is a normal to the ellipse \( 3x^2 + 8y^2 = k \), then the equation of the tangent at point (-2,m) is:
If \( 4x - 3y - 5 = 0 \) is a normal to the ellipse \( 3x^2 + 8y^2 = k \), then the equation of the tangent at point (-2,m) is:
Show Hint
For ellipse normal and tangent problems, use implicit differentiation or standard normal-tangent relations.
Updated On: Apr 18, 2025
- \( 3x + 4y - 14 = 0 \)
- \( 3x - 4y + 10 = 0 \)
- \( 3x - 4y + 1 = 0 \)
- \( 4x + 3y - 3 = 0 \)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Step 1: Using the normal equation condition
A normal to an ellipse satisfies the equation:
\[
ax + by + c = 0
\]
We substitute \( x = -2 \), solve for \( y \), and derive the tangent equation.
Was this answer helpful?
0
1
Top AP EAMCET Mathematics Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
- If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is a focal chord, then one possible values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is
View More Questions
Top AP EAMCET Ellipse Questions
- If the line joining the points $A(\alpha)$ and $B(\beta)$ on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ is a focal chord, then one possible values of $\cot \frac{\alpha}{2} . \cot \frac{\beta}{2}$ is
- $P$ is a variable point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $ with foci $F_1$ and $F_2$ . If $A$ is the area of the triangle $PF_1F_2$. then the maximum value of $A$ is
- Let \( T_1 \) be the tangent drawn at a point \( P(\sqrt{2}, \sqrt{3}) \) on the ellipse \( \frac{x^2}{4} + \frac{y^2}{6} = 1 \). If \( (a, \beta) \) is the point where \( T_1 \) intersects another tangent \( T_2 \) to the ellipse perpendicularly, then \( a^2 + \beta^2 = \):
- The product of perpendiculars from the two foci of the ellipse \[ \frac{x^2}{9} + \frac{y^2}{25} = 1 \] on the tangent at any point on the ellipse is:
If a tangent of slope 2 to the ellipse \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) touches the circle \(x^2 + y^2 = 4\), then the maximum value of ab is:
Top AP EAMCET Questions
- If $\int \cos x . \cos 2x . \cos 5x dx = A \; \sin 2x + B \sin 4x + C \sin 6x + D \sin 8x + k $ (where $k$ is the arbitrary constant of integration), then $\frac{1}{B} + \frac{1}{C} = $
- If $k$ is one of the roots of the equation $x^2 - 25x + 24 = 0 $ such that $A = \begin{bmatrix}1&2&1\\ 3&2&3\\ 1&1&k\end{bmatrix} $ is a non-singular matrix, then $A^{-1}$ =
- If only $\frac{1}{51}^{th}$ ot the main current is to be passed through a galvanometer then the shunt required is $R_1$ and if only $\frac{1}{11}^{th}$ of the main voltage is to be developed across the galvanometer, then the resistance required is $R_2$. Then $\frac{R_2}{R_1} =$
- If $p$ and $q$ are respectively the global maximum and global minimum of the function $f(x) = x^2 e^{2x}$ on the interval $[-2, 2]$ , then $pe^{-4} + qe^4 = $
- If $\displaystyle\sum^n_{k - 1} \tan^{-1} \left( \frac{1}{k^2 + k + 1} \right) = \tan^{-1} (\theta) $ , then $\theta$ =
View More Questions