Two common tangents to the circle {x^2} + {y^ | vedclass

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

Question

(B) The equation of any tangent to the parabola ${y^2} = 8ax$ is given by $y = mx + \frac{2a}{m}$.This line is also a tangent to the circle ${x^2} + {

Vedclass · Accepted Answer

$y = \pm (x + 2a)$

Vedclass · Answer

(B) The equation of any tangent to the parabola ${y^2} = 8ax$ is given by $y = mx + \frac{2a}{m}$.This line is also a tangent to the circle ${x^2} + {y^2} = 2{a^2}$.The perpendicular distance from the center $(0, 0)$ of the circle to the line $mx - y + \frac{2a}{m} = 0$ must be equal to the radius $r = \sqrt{2}a$.Thus,$\frac{|\frac{2a}{m}|}{\sqrt{m^2 + 1}} = \sqrt{2}a$.Squaring both sides,we get $\frac{4a^2}{m^2(m^2 + 1)} = 2a^2$.This simplifies to $m^2(m^2 + 1) = 2$,or $m^4 + m^2 - 2 = 0$.Factoring the quadratic in $m^2$,we have $(m^2 - 1)(m^2 + 2) = 0$.Since $m^2$ must be real and positive,we have $m^2 = 1$,which gives $m = \pm 1$.Substituting $m = \pm 1$ into the tangent equation $y = mx + \frac{2a}{m}$,we get $y = \pm (x + 2a)$.

Two common tangents to the circle ${x^2} + {y^2} = 2{a^2}$ and the parabola ${y^2} = 8ax$ are

Similar Questions

The area of the region (in sq. unit) bounded by the $x$-axis, the tangent, and the normal to the circle $x^2 + y^2 = 4$ drawn at the point $(1, \sqrt{3})$ is

The angle between the two tangents drawn from the origin to the circle $x^2+y^2-14x+2y+25=0$ is (in $^{\circ}$)

The tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$. Find its point of contact.

The point of contact of the tangent to the circle $x^2 + y^2 = 5$ at the point $(1, -2)$ which also touches the circle $x^2 + y^2 - 8x + 6y + 20 = 0$ is:

The equations of the tangents to the circle $x^2+y^2=4$ drawn from the point $(4,0)$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module