Let P(asec theta, btan theta) and Q(asec varp | vedclass

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The equation of the normal at $P(a\sec 	heta, b	an 	heta)$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $ax\sin 	heta + by = (a^2 + b^2)	an 	heta$. Similarly,the equation of the normal at $Q(a\sec \varphi, b	an \varphi)$ is $ax\sin \varphi + by = (a^2 + b^2)	an \varphi$. Let $(h, k)$ be the intersection point. Then $k$ satisfies: $by = (a^2 + b^2)	an 	heta - ax\sin 	heta$ and $by = (a^2 + b^2)	an \varphi - ax\sin \varphi$. Subtracting the two equations: $b(k - k) = 0$ is not the path; we solve for $y$ by eliminating $x$. From the normal equations: $ax\sin 	heta + by = (a^2 + b^2)	an 	heta$ and $ax\sin \varphi + by = (a^2 + b^2)	an \varphi$. Multiplying the first by $\sin \varphi$ and the second by $\sin 	heta$ and subtracting: $by(\sin \varphi - \sin 	heta) = (a^2 + b^2)(	an 	heta \sin \varphi - 	an \varphi \sin 	heta)$. Given $\varphi = \frac{\pi}{2} - 	heta$,so $\sin \varphi = \cos 	heta$ and $	an \varphi = \cot 	heta$. $by(\cos 	heta - \sin 	heta) = (a^2 + b^2)(	an 	heta \cos 	heta - \cot 	heta \sin 	heta)$. $by(\cos 	heta - \sin 	heta) = (a^2 + b^2)(\sin 	heta - \cos 	heta)$. $by = -(a^2 + b^2)$. Thus,$k = -\frac{a^2 + b^2}{b}$.

Let $P(a\sec \theta, b\tan \theta)$ and $Q(a\sec \varphi, b\tan \varphi)$,where $\theta + \varphi = \frac{\pi}{2}$,be two points 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 $k$ is equal to

Similar Questions

Water flows in a streamlined manner through a capillary tube of radius $a$,the pressure difference being $P$ and the rate of flow $Q$. If the radius is reduced to $a/2$ and the pressure is increased to $2P$,the rate of flow becomes:

If the length of a stretched string is shortened by $40\%$ and the tension is increased by $44\%$,then the ratio of the final and initial fundamental frequencies is

If $a_k$ is the coefficient of $x^k$ in the expansion of $(1+x+x^2)^n$ for $k=0, 1, 2, \ldots, 2n$,then $a_1+2a_2+3a_3+\ldots+2na_{2n}$ is equal to

If $l, m, n$ are in arithmetic progression,then the straight line $lx + my + n = 0$ will always pass through the point:

Vedclass Test Series

Exam Paper Generator

Online Exam Module