If tan theta_1 cdot tan theta_2 = -frac{a^2}{ | vedclass

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

Question

(B) Let the points on the ellipse be $P(a \cos 	heta_1, b \sin 	heta_1)$ and $Q(a \cos 	heta_2, b \sin 	heta_2)$.The slopes of the lines joining t

Vedclass · Accepted Answer

centre

Vedclass · Answer

(B) Let the points on the ellipse be $P(a \cos 	heta_1, b \sin 	heta_1)$ and $Q(a \cos 	heta_2, b \sin 	heta_2)$.The slopes of the lines joining the origin $O(0,0)$ to $P$ and $Q$ are $m_1 = \frac{b \sin 	heta_1}{a \cos 	heta_1} = \frac{b}{a} 	an 	heta_1$ and $m_2 = \frac{b \sin 	heta_2}{a \cos 	heta_2} = \frac{b}{a} 	an 	heta_2$.The product of the slopes is $m_1 m_2 = \frac{b^2}{a^2} 	an 	heta_1 	an 	heta_2$.Given $	an 	heta_1 	an 	heta_2 = -\frac{a^2}{b^2}$,we have $m_1 m_2 = \frac{b^2}{a^2} \left(-\frac{a^2}{b^2}ight) = -1$.Since the product of the slopes is $-1$,the lines $OP$ and $OQ$ are perpendicular,meaning the chord $PQ$ subtends a right angle at the centre $O(0,0)$.

If $\tan \theta_1 \cdot \tan \theta_2 = -\frac{a^2}{b^2}$,then the chord joining two points $\theta_1$ and $\theta_2$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ will subtend a right angle at:

Similar Questions

The center of the ellipse $x^2+2y^2-4x+12y+14=0$ is

If the latus rectum of an ellipse is equal to half of its minor axis,then its eccentricity is

The value of $k$,if $(1, 2)$ and $(k, -1)$ are conjugate points with respect to the ellipse $2x^2 + 3y^2 = 6$,is

$A$ line of fixed length $a + b$, where $a \neq b$, moves such that its ends are always on two fixed perpendicular straight lines. The locus of a point which divides the line into two parts of lengths $a$ and $b$ is

If $P$ is any point on the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$ and $S, S^{\prime}$ are its foci,then the maximum area (in sq. units) of $\Delta SPS^{\prime} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module