Question:
A normal is drawn at a point $(x_1, y_1)$ of the parabola $y^2 = 16x$ and this normal makes equal angle with both $x$ and $y$ axes. Then point $(x_1, y_1)$ is
A normal is drawn at a point $(x_1, y_1)$ of the parabola $y^2 = 16x$ and this normal makes equal angle with both $x$ and $y$ axes. Then point $(x_1, y_1)$ is
Updated On: Jun 20, 2022
- $(4, - 4)$
- $(2, - 8)$
- $(4, - 8)$
- $(1, - 4)$
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The Correct Option is C
Solution and Explanation
Given equation of parabola is
$y^{2}=16 x$
On differentiating both sides, we get
$2 y y'=16$
$y'=\frac{16}{2 y}=\frac{8}{y}$
$\therefore$ Slope of tangent at point $\left(x_{1}, y_{1}\right), m_{1}=\frac{8}{y_{1}}$
and slope of normal at point $\left(x_{1}, y_{1}\right), m_{2}=\frac{-y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ -axes, then
$m_{2}=\pm 1$
$\Rightarrow \frac{-y_{1}}{8}=\pm 1$
$\Rightarrow -y_{1}=\pm 8$
Now, when $y_{1}=8$, then $x_{1}=4$
when $y_{1}=-8$, then $x_{1}=4$
So, the required point is $(4,-8)$
$y^{2}=16 x$
On differentiating both sides, we get
$2 y y'=16$
$y'=\frac{16}{2 y}=\frac{8}{y}$
$\therefore$ Slope of tangent at point $\left(x_{1}, y_{1}\right), m_{1}=\frac{8}{y_{1}}$
and slope of normal at point $\left(x_{1}, y_{1}\right), m_{2}=\frac{-y_{1}}{8}$
Since, normal makes equal angle with both $X$ and $Y$ -axes, then
$m_{2}=\pm 1$
$\Rightarrow \frac{-y_{1}}{8}=\pm 1$
$\Rightarrow -y_{1}=\pm 8$
Now, when $y_{1}=8$, then $x_{1}=4$
when $y_{1}=-8$, then $x_{1}=4$
So, the required point is $(4,-8)$
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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.