An ellipse has eccentricity frac{1}{2} and on | vedclass

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

Question

(A) The common tangents to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$ are $x = 1$ and $x = -1$.Out of these, $x = 1$ is nearer to th

Vedclass · Accepted Answer

$\frac{(x - 1/3)^2}{1/9} + \frac{(y - 1)^2}{1/12} = 1$

Vedclass · Answer

(A) The common tangents to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$ are $x = 1$ and $x = -1$.Out of these, $x = 1$ is nearer to the point $P\left( \frac{1}{2}, 1 ight)$.Thus, the directrix of the required ellipse is $x = 1$.Let $Q(x, y)$ be any point on the ellipse.By the definition of an ellipse, the distance from the focus $P$ is $QP = e 	imes (	ext{distance from directrix})$.$QP = \sqrt{(x - 1/2)^2 + (y - 1)^2}$ and the distance from the directrix $x = 1$ is $|x - 1|$.Given $e = 1/2$, we have $\sqrt{(x - 1/2)^2 + (y - 1)^2} = \frac{1}{2} |x - 1|$.Squaring both sides: $(x - 1/2)^2 + (y - 1)^2 = \frac{1}{4}(x - 1)^2$.$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}(x^2 - 2x + 1)$.$x^2 - x + 1/4 + (y - 1)^2 = \frac{1}{4}x^2 - \frac{1}{2}x + 1/4$.$\frac{3}{4}x^2 - \frac{1}{2}x + (y - 1)^2 = 0$.Multiply by $4/3$: $x^2 - \frac{2}{3}x + \frac{4}{3}(y - 1)^2 = 0$.$(x^2 - \frac{2}{3}x + 1/9) + \frac{4}{3}(y - 1)^2 = 1/9$.$(x - 1/3)^2 + \frac{4}{3}(y - 1)^2 = 1/9$.Dividing by $1/9$: $\frac{(x - 1/3)^2}{1/9} + \frac{(y - 1)^2}{1/12} = 1$.

$Column 1$	$Column 2$	$Column 3$
$(I) x^2+y^2=a^2$	$(i) my=m^2x+a$	$(P) (a/m^2, 2a/m)$
$(II) x^2+a^2y^2=a^2$	$(ii) y=mx+a\sqrt{m^2+1}$	$(Q) (-ma/\sqrt{m^2+1}, a/\sqrt{m^2+1})$
$(III) y^2=4ax$	$(iii) y=mx+\sqrt{a^2m^2-1}$	$(R) (-a^2m/\sqrt{a^2m^2+1}, 1/\sqrt{a^2m^2+1})$
$(IV) x^2-a^2y^2=a^2$	$(iv) y=mx+\sqrt{a^2m^2+1}$	$(S) (-a^2m/\sqrt{a^2m^2-1}, -1/\sqrt{a^2m^2-1})$

An ellipse has eccentricity $\frac{1}{2}$ and one focus at the point $P\left( \frac{1}{2}, 1 \right)$. Its one directrix is the common tangent nearer to the point $P$, to the circle $x^2 + y^2 = 1$ and the hyperbola $x^2 - y^2 = 1$. The equation of the ellipse in the standard form is:

Similar Questions

The equation of the line passing through the points of intersection of the parabola $x^2 = 8y$ and the ellipse $\frac{x^2}{3} + y^2 = 1$ is

The locus of the foot of the perpendicular drawn from the centre upon any tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

The point$(s)$ on the parabola $y^2 = 4x$ which are closest to the circle $x^2 + y^2 - 24y + 128 = 0$ is/are:

If the normal to the ellipse $\frac{x^2}{16} + \frac{y^2}{3} = 1$ at the point $\left( 2, \frac{3}{2} \right)$ is tangent to a parabola,find the equation of the parabola.

Vedclass Test Series

Exam Paper Generator

Online Exam Module