Let P(a sec theta, b tan theta) and Q(a sec p | vedclass

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

Question

(D) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.The equation of the normal at a point $P(a \sec 	heta, b 	an 	heta)$ is $\

Vedclass · Accepted Answer

$-\left(\frac{a^2+b^2}{b}\right)$

Vedclass · Answer

(D) The equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$.The equation of the normal at a point $P(a \sec 	heta, b 	an 	heta)$ is $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2+b^2$,which simplifies to $ax \cos 	heta + by \cot 	heta = a^2+b^2$.Similarly,the equation of the normal at $Q(a \sec \phi, b 	an \phi)$ is $ax \cos \phi + by \cot \phi = a^2+b^2$.Given $	heta + \phi = \frac{\pi}{2}$,we have $\phi = \frac{\pi}{2} - 	heta$. Thus,$\cos \phi = \sin 	heta$ and $\cot \phi = 	an 	heta$.The two equations are:$(1) \quad ax \cos 	heta + by \cot 	heta = a^2+b^2$$(2) \quad ax \sin 	heta + by 	an 	heta = a^2+b^2$To find $k$ (the $y$-coordinate),we eliminate $x$ by multiplying $(1)$ by $\sin 	heta$ and $(2)$ by $\cos 	heta$:$ax \cos 	heta \sin 	heta + by \cot 	heta \sin 	heta = (a^2+b^2) \sin 	heta$$ax \sin 	heta \cos 	heta + by 	an 	heta \cos 	heta = (a^2+b^2) \cos 	heta$Subtracting the two equations:$by(\cot 	heta \sin 	heta - 	an 	heta \cos 	heta) = (a^2+b^2)(\sin 	heta - \cos 	heta)$$by(\cos 	heta - \sin 	heta) = (a^2+b^2)(\sin 	heta - \cos 	heta)$$by = -(a^2+b^2)$$y = -\frac{a^2+b^2}{b}$Therefore,$k = -\frac{a^2+b^2}{b}$.

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If $(h, k)$ is the point of intersection of the normals at $P$ and $Q$,then $k=$

Similar Questions

The eccentricity of the conic ${x^2} - 4{y^2} = 1$ is

If the tangent at the point $(2 \sec \phi, 3 \tan \phi)$ of the hyperbola $\frac{x^2}{4} - \frac{y^2}{9} = 1$ is parallel to $3x - y + 4 = 0$,then the value of $\phi$ is ............ $^o$.

If $e_1$ and $e_2$ are respectively the eccentricities of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and its conjugate hyperbola,then the line $\frac{x}{2 e_1}+\frac{y}{2 e_2}=1$ touches the circle having centre at the origin. Find its radius.

If the ratio of the distance between the foci to the distance between the directrices of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $3 : 2$,then $a : b = \dots$

The length of the latus rectum of the hyperbola $\frac{x^2}{\cos^2 \alpha} - \frac{y^2}{\sin^2 \alpha} = 4$ is (where $\alpha \neq \frac{n\pi}{2}, n \in I$).

Vedclass Test Series

Exam Paper Generator

Online Exam Module