If the curve x^{2}+2 y^{2}=2 intersects the l | vedclass

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

Question

(D) To find the angle subtended by the chord $PQ$ at the origin,we homogenize the equation of the curve $x^{2} + 2y^{2} = 2$ using the line $x + y = 1

Vedclass · Accepted Answer

$\frac{\pi}{2}+	an ^{-1}\left(\frac{1}{4}ight)$

Vedclass · Answer

(D) To find the angle subtended by the chord $PQ$ at the origin,we homogenize the equation of the curve $x^{2} + 2y^{2} = 2$ using the line $x + y = 1$.The equation of the line can be written as $x + y = 1$,so $1 = x + y$.Substituting this into the curve equation:$x^{2} + 2y^{2} = 2(1)^{2}$$x^{2} + 2y^{2} = 2(x + y)^{2}$$x^{2} + 2y^{2} = 2(x^{2} + 2xy + y^{2})$$x^{2} + 2y^{2} = 2x^{2} + 4xy + 2y^{2}$$x^{2} + 4xy = 0$This represents a pair of lines passing through the origin. Let these lines be $y = m_{1}x$ and $y = m_{2}x$.From $x(x + 4y) = 0$,we get $x = 0$ (slope $m_{1} = \infty$) and $y = -\frac{1}{4}x$ (slope $m_{2} = -\frac{1}{4}$).The angle $	heta$ between these two lines is given by:$	an 	heta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} ight|$Since one line is vertical $(x=0)$,the angle $	heta$ is the angle between the vertical line and the line with slope $-\frac{1}{4}$.The angle of the line $y = -\frac{1}{4}x$ with the $x$-axis is $\alpha = 	an^{-1}(-\frac{1}{4}) = -	an^{-1}(\frac{1}{4})$.The angle between the vertical line $(90^{\circ})$ and this line is $\frac{\pi}{2} - (-	an^{-1}(\frac{1}{4})) = \frac{\pi}{2} + 	an^{-1}(\frac{1}{4})$.

If the curve $x^{2}+2 y^{2}=2$ intersects the line $x + y =1$ at two points $P$ and $Q$,then the angle subtended by the line segment $PQ$ at the origin is ...... .

Similar Questions

The quadratic equation whose roots are $l$ and $m$,where $l = \lim_{\theta \rightarrow 0} \left( \frac{3 \sin \theta - 4 \sin^2 \theta}{\theta} \right)$ and $m = \lim_{\theta \rightarrow 0} \frac{2 \tan \theta}{\theta(1 - \tan^2 \theta)}$,is:

Find the angle of intersection of the curves $y^{2}=4ax$ and $x^{2}=4by$.

If the curves $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and $x^2 - y^2 = c^2$ intersect at right angles,then:

If $e$ and $e^{\prime}$ are the eccentricities of the ellipse $5x^2 + 9y^2 = 45$ and the hyperbola $5x^2 - 4y^2 = 45$ respectively,then $ee^{\prime}$ is equal to

The point$(s)$ on the parabola $y^2 = 4x$ which are closest to the circle $x^2 + y^2 - 24y + 128 = 0$ is/are:

Vedclass Test Series

Exam Paper Generator

Online Exam Module