Find the equations of the tangent and normal | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$,we get $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$. Thus,$\

Vedclass · Accepted Answer

Tangent: $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$,Normal: $\frac{y-y_{0}}{a^{2} y_{0}}+\frac{x-x_{0}}{b^{2} x_{0}}=0$

Vedclass · Answer

(D) Differentiating $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ with respect to $x$,we get $\frac{2x}{a^{2}}-\frac{2y}{b^{2}}\frac{dy}{dx}=0$. Thus,$\frac{dy}{dx}=\frac{b^{2}x}{a^{2}y}$. The slope of the tangent at $(x_{0}, y_{0})$ is $m_{t} = \frac{b^{2}x_{0}}{a^{2}y_{0}}$. The equation of the tangent is $y-y_{0} = \frac{b^{2}x_{0}}{a^{2}y_{0}}(x-x_{0})$,which simplifies to $\frac{x x_{0}}{a^{2}}-\frac{y y_{0}}{b^{2}}=1$. The slope of the normal is $m_{n} = -\frac{1}{m_{t}} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}$. The equation of the normal is $y-y_{0} = -\frac{a^{2}y_{0}}{b^{2}x_{0}}(x-x_{0})$,which can be written as $\frac{y-y_{0}}{a^{2}y_{0}} + \frac{x-x_{0}}{b^{2}x_{0}} = 0$.

Find the equations of the tangent and normal to the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at the point $(x_{0}, y_{0})$.

Similar Questions

The locus of a point which moves such that the difference of its distances from two fixed points is always a constant is

If the latus rectum of a hyperbola through one focus subtends an angle of $60^{\circ}$ at the other focus,then its eccentricity is

Tangent at any point $\theta$ on the curve $x=35 \sec \theta, y=35 \tan \theta$ is

The locus of a point whose chord of contact with respect to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ touches the circle described on the straight line joining the foci of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ as diameter is

The number of integral points $(x, y)$ interior to the circle $x^2 + y^2 = 10$ from which exactly one real tangent can be drawn to the curve $\sqrt{(x + 5\sqrt{2})^2 + y^2} - \sqrt{(x - 5\sqrt{2})^2 + y^2} = 10$ is (where an integral point $(x, y)$ means $x, y \in \mathbb{Z}$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module