The lines y=2x+sqrt{76} and 2y+x=8 touch the | vedclass

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

Question

(A) The given equations of the lines are $y=2x+\sqrt{76}$ and $y=-\frac{1}{2}x+4$. Comparing these with the slope-intercept form $y=mx+c$,we get slope

Vedclass · Accepted Answer

$x^2+y^2=28$

Vedclass · Answer

(A) The given equations of the lines are $y=2x+\sqrt{76}$ and $y=-\frac{1}{2}x+4$. Comparing these with the slope-intercept form $y=mx+c$,we get slopes $m_1=2$ and $m_2=-\frac{1}{2}$. Since $m_1 	imes m_2 = 2 	imes (-\frac{1}{2}) = -1$,the two lines are perpendicular to each other. The locus of the point of intersection of two perpendicular tangents to an ellipse is its director circle. The equation of the director circle for an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is $x^2+y^2=a^2+b^2$. Here,$a^2=16$ and $b^2=12$. Therefore,the equation of the circle is $x^2+y^2=16+12=28$.

The lines $y=2x+\sqrt{76}$ and $2y+x=8$ touch the ellipse $\frac{x^2}{16}+\frac{y^2}{12}=1$. If the point of intersection of these two lines lies on a circle whose centre coincides with the centre of that ellipse,then the equation of that circle is

Similar Questions

If a tangent with slope $-4/3$ to the ellipse $\frac{x^2}{18} + \frac{y^2}{32} = 1$ intersects the major axis and minor axis at $A$ and $B$ respectively,then the area of $\Delta OAB$ is .......... square units.

An ellipse has its center at $(1, -2)$, one focus at $(3, -2)$, and one vertex at $(5, -2)$. Then the length of its latus rectum is :

Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2\sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis,then $SP^{2} + SQ^{2}$ is equal to.

If $S$ and $S^{\prime}$ are the foci of an ellipse,$B$ is one end of the minor axis and $\angle SBS^{\prime} = 90^{\circ}$,then the eccentricity of that ellipse is

Find the length of the latus rectum of the ellipse $4x^2 + 9y^2 - 36y + 4 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module