Question:
The equation to the normal to the hyperbola $\frac {x^2}{16}- \frac {y^2}{9}=1$ at $(-4,0)$ is
The equation to the normal to the hyperbola $\frac {x^2}{16}- \frac {y^2}{9}=1$ at $(-4,0)$ is
Updated On: Apr 12, 2024
- 2x - 3y = 1
- x = 0
- x = 1
- y = 0
Hide Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
We know that, the equation of normal at the
point $\left(x_{1}, y_{1}\right)$ to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is $\frac{a^{2} x}{x_{1}}+\frac{b^{2} y}{y_{1}}=a^{2}+b^{2}$
Given equation is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Here, $\,\,\,\,\,\,a^{2}=16,\,\, b^{2}=9$
$\therefore\,\,\,$ The equation of normal at the point $(-4,0)$ is
$\frac{16 x}{-4}+\frac{9 y}{0}=16+9$
$\Rightarrow\,\,\,\,\, \frac{9 y}{0}=25+\frac{16 x}{4}$
$\Rightarrow \,\,\,\,\,\,9 y=0 $
$ \Rightarrow \,\,\,\,y=0$
Was this answer helpful?
1
0
Top Questions on Hyperbola
- 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 ________.
- Consider a hyperbola \( H \) having its centre at the origin and foci on the \( x \)-axis. Let \( C_1 \) be the circle touching the hyperbola \( H \) and having its centre at the origin. Let \( C_2 \) be the circle touching the hyperbola \( H \) at its vertex and having its centre at one of its foci. If the areas (in square units) of \( C_1 \) and \( C_2 \) are \( 36\pi \) and \( 4\pi \), respectively, then the length (in units) of the latus rectum of \( H \) is:
- Let H: $\frac{-x^2}{a^2} + \frac{y^2}{b^2} = 1$ be the hyperbola, whose eccentricity is $\sqrt{3}$ and the length of the latus rectum is $4\sqrt{3}$. Suppose the point $(\alpha, 6)$, $\alpha>0$ lies on H. If $\beta$ is the product of the focal distances of the point $(\alpha, 6)$, then $\alpha^2 + \beta$ is equal to:
- If the hyperbola \(x² - y² cosec²θ = 5\) and ellipse \(x²cosec²θ + y² = 5\) has eccentricity \(e_H\) and \(e_E\) respectively and \(e_H = \sqrt 7e_E\), then \(θ\) is equal to
- The foci of a hyperbola are (±2,0) and its eccentricity is \(\frac{3}{2}\) . A tangent, perpendicular to the line 2x + 3y = 6, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the x- and y-axes are a and b respectively, then |6a| + |5b| is equal to_____.
View More Questions
Questions Asked in KCET exam
- Ten identical cells each emf 2V and internal resistance 1Ω are connected in series with two cells wrongly connected. A resistor of 10Ω is connected to the combination. What is the current through the resistor?
- KCET - 2023
- Cells in Series and in Parallel
- The speed of sound in an ideal gas at a given temperature T is v. The rms speed of gas molecules at that temperature is vrms. The ratio of the velocities v and vrms for helium and oxygen gases are X and X' respectively. Then \(\frac{X}{X'}\) is equal to
- KCET - 2023
- Thermodynamics
- If \(p(\frac{1}{q}+\frac{1}{r}),q(\frac{1}{r}+\frac{1}{p}),r(\frac{1}{p}+\frac{1}{q})\) are in A.P., then p, q, r
- KCET - 2023
- Arithmetic Progression
- A stretched wire of a material whose Young's modulus Y = 2 × 1011 Nm-2 has Poisson's ratio of 0.25. Its lateral strain εl = 10-3. The elastic energy density of the wire is
- KCET - 2023
- mechanical properties of solids
- A metallic rod of length 1 m held along east-west direction is allowed to fall down freely. Given horizontal component of earth’s magnetic field BH = 3 × 10-5 T. The emf induced in the rod at an instant t = 2s after it is released is ( Take g = 10 ms-2 )
- KCET - 2023
- Faradays laws of induction
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.
