Equations of common tangents to ellipse, x^2 + 4y^2 = 8 & parabola y^2 = 4x can be

English

Gujarati

Hindi

10-2. Parabola, Ellipse, Hyperbola

normal

The equations of the common tangents to the ellipse, $ x^2 + 4y^2 = 8 $ $\&$ the parabola $y^2 = 4x$ can be

A

$x + 2y + 4 = 0$

B

$x - 2y + 4 = 0$

C

$2x + y - 4 = 0$

D

both $(A)$ and $(B)$

Solution

Equation of tangent to above curves are respectively.

$y ^{2}= mx +\frac{1}{ m }$ and $y = mx +\sqrt{8 m ^{2}+2}$

Comparing $\frac{1}{ m }=\sqrt{8 m ^{2}+2}$

$\Rightarrow m ^{2}\left(8 m ^{2}+2\right)=1$

seeing the options

$m =\pm \frac{1}{2}$ satisfy the equation

$\Rightarrow y=\pm \frac{1}{2} x \pm 2 \Rightarrow 2 y=\pm x \pm 4$

i.e. $2 y=x+4 \& x+2 y+4=0$

Standard 11

Mathematics

Share

Similar Questions

If any tangent to the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ cuts off intercepts of length $h$ and $k$ on the axes, then $\frac{{{a^2}}}{{{h^2}}} + \frac{{{b^2}}}{{{k^2}}} = $

hard

If $P$ lies in the first quadrant on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$ (where $a > b$ ), and tangent & normal drawn at $P$ meets major axis at the points $T$ & $N$ respectively, then the value of $\frac{{\left( {\left| {{F_2}N} \right| + \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| – \left| {{F_1}T} \right|} \right)}}{{\left( {\left| {{F_2}N} \right| – \left| {{F_1}N} \right|} \right)\left( {\left| {{F_2}T} \right| + \left| {{F_1}T} \right|} \right)}}$ is equal to (where $F_1$ & $F_2$ are the foci $(ae, 0)$ & $(-ae, 0)$ respectively)

normal

If $ \tan\ \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$ then the chord joining two points $\theta _1 \& \theta _2$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$ will subtend a right angle at :

normal

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

normal

(KVPY-2017)

The foci of the ellipse $25{(x + 1)^2} + 9{(y + 2)^2} = 225$ are at

hard