Question:
The normal at $\left(2,\frac{3}{2}\right)$ to the ellipse, $\frac{x^{2}}{16}+\frac{y^{2}}{3} = 1$ touches a parabola, whose equation is
The normal at $\left(2,\frac{3}{2}\right)$ to the ellipse, $\frac{x^{2}}{16}+\frac{y^{2}}{3} = 1$ touches a parabola, whose equation is
Updated On: Jan 29, 2024
- $y^2 =- 10^4\,x$
- $y^2=14\,x$
- $y^2 = 26\,x$
- $y^2 = - 14\,x$
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The Correct Option is A
Solution and Explanation
Ellipse is $ \frac{x^{2}}{16}+\frac{y^{2}}{3} = 1$
Now, equation of normal at $\left(2, 3/2\right)$ is
$\frac{16x}{2}-\frac{3y}{3/2} = 16-3$
$\Rightarrow\quad8x-2y= 13$
$\Rightarrow \quad y=4x-\frac{13}{2}$
Let $y=4x-\frac{13}{2}$ touches a parabola
$y^{2} = 4ax.$
We know, a straight liney $= mx + c$ touches a parabola $y^{2}=4ax$ if a-me $= 0$
$\therefore\quad a-\left(4\right)\left(-\frac{13}{2}\right) = 0 \Rightarrow a = -26$
Hence, required equation of parabola is
$y^{2} = 4\left(-26\right)x = -104\,x$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.