If a tangent having slope 2 of the ellipse (x2/a2)+(y2/b2)=1 is normal to the circle x2+y2+4x+1=0, then the value of 4a2+b2 is equal to

Question

If a tangent having slope 2 of the ellipse (x2/a2)+(y2/b2)=1 is normal to the circle x2+y2+4x+1=0, then the value of 4a2+b2 is equal to (A) 4 (B) 2 (C) 16 (D) 8

Tardigrade Admin · Accepted Answer

Equation of tangent is y=2x±√4 a2 + b2 This is normal to the circle x2+y2+4x+1=0 This tangent passes through (- 2,0) 0=-4±√4 a2 + b2⇒ 4a2+b2=16

Q. If a tangent having slope $2$ of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0,$ then the value of $4 a^{2} + b^{2}$ is equal to

$4$

$2$

$16$

$8$

Equation of tangent is $y = 2 x \pm 4 a^{2} + b^{2}$
This is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0$
This tangent passes through $(- 2, 0)$
$0 = - 4 \pm 4 a^{2} + b^{2} \Rightarrow 4 a^{2} + b^{2} = 16$

Q. If a tangent having slope 2 of the ellipse a2x2​+b2y2​=1 is normal to the circle x2+y2+4x+1=0, then the value of 4a2+b2 is equal to

4

2

16

8

Equation of tangent is y=2x±4a2+b2​ This is normal to the circle x2+y2+4x+1=0 This tangent passes through (−2,0) 0=−4±4a2+b2​⇒4a2+b2=16

Q. If a tangent having slope $2$ of the ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0,$ then the value of $4 a^{2} + b^{2}$ is equal to

$4$

$2$

$16$

$8$

Equation of tangent is $y = 2 x \pm 4 a^{2} + b^{2}$
This is normal to the circle $x^{2} + y^{2} + 4 x + 1 = 0$
This tangent passes through $(- 2, 0)$
$0 = - 4 \pm 4 a^{2} + b^{2} \Rightarrow 4 a^{2} + b^{2} = 16$