If the common tangents to the parabola x^2 = | vedclass

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

Question

(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.Since this line is also a tangent to the pa

Vedclass · Accepted Answer

$3 + 2\sqrt{2}$

Vedclass · Answer

(D) The equation of a tangent to the circle $x^2 + y^2 = 4$ with slope $m$ is $y = mx \pm 2\sqrt{1 + m^2}$.Since this line is also a tangent to the parabola $x^2 = 4y$,we substitute $y = mx + c$ into the parabola equation: $x^2 = 4(mx + c) \Rightarrow x^2 - 4mx - 4c = 0$.For tangency,the discriminant $D = 0$,so $(-4m)^2 - 4(1)(-4c) = 0$ $\Rightarrow 16m^2 + 16c = 0$ $\Rightarrow c = -m^2$.Comparing this with the circle's tangent form $c = \pm 2\sqrt{1 + m^2}$,we get $-m^2 = \pm 2\sqrt{1 + m^2}$.Squaring both sides: $m^4 = 4(1 + m^2) \Rightarrow m^4 - 4m^2 - 4 = 0$.Using the quadratic formula for $m^2$: $m^2 = \frac{4 \pm \sqrt{16 - 4(1)(-4)}}{2} = \frac{4 \pm \sqrt{32}}{2} = 2 \pm 2\sqrt{2}$.Since $m^2$ must be positive,$m^2 = 2 + 2\sqrt{2}$.

If the common tangents to the parabola $x^2 = 4y$ and the circle $x^2 + y^2 = 4$ intersect at the point $P$,then find the square of the slope of the line.

Similar Questions

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(b>a)$ and the parabola $y^2=8ax$ intersect at right angles. If $e$ is the eccentricity of the ellipse,then $e^4$ is equal to

If for $\theta \in \left[-\frac{\pi}{3}, 0\right]$,the points $(x, y) = \left(3 \tan \left(\theta+\frac{\pi}{3}\right), 2 \tan \left(\theta+\frac{\pi}{6}\right)\right)$ lie on $xy+\alpha x+\beta y+\gamma=0$,then $\alpha^2+\beta^2+\gamma^2$ is equal to:

Find the locus of a point which moves such that its distance from the point $(0, 0)$ is twice its distance from the $y$-axis.

The locus of the midpoints of the chords of the hyperbola $x^2 - y^2 = a^2$ which are tangents to the parabola $x^2 = 4by$ will be -

The equation of the common tangent to the curves $y^2 = 8x$ and $xy = -1$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module