The angle subtended by the normal chord at th | vedclass

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

Question

(C) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.The focus of the parabola is $S = (a,

Vedclass · Accepted Answer

$90$

Vedclass · Answer

(C) The equation of the parabola is $y^2 = 9x$. Comparing with $y^2 = 4ax$,we get $4a = 9$,so $a = \frac{9}{4}$.The focus of the parabola is $S = (a, 0) = (\frac{9}{4}, 0)$.The point $P$ is $(9, 9)$. Let the normal at $P$ meet the parabola again at $Q$.The slope of the tangent at $P(x_1, y_1)$ is $m = \frac{2a}{y_1} = \frac{9/2}{9} = \frac{1}{2}$.The slope of the normal at $P$ is $m' = -2$.The equation of the normal at $P(9, 9)$ is $y - 9 = -2(x - 9)$,which simplifies to $y = -2x + 27$.Substituting $y = -2x + 27$ into $y^2 = 9x$ gives $(-2x + 27)^2 = 9x$,which is $4x^2 - 108x + 729 = 9x$,or $4x^2 - 117x + 729 = 0$.Since $x = 9$ is one root,the other root $x_2$ satisfies $9 	imes x_2 = \frac{729}{4}$,so $x_2 = \frac{81}{4}$.Then $y_2 = -2(\frac{81}{4}) + 27 = -\frac{81}{2} + \frac{54}{2} = -\frac{27}{2}$.The focus is $S = (\frac{9}{4}, 0)$. The slopes of $SP$ and $SQ$ are $m_1 = \frac{9 - 0}{9 - 9/4} = \frac{9}{27/4} = \frac{4}{3}$ and $m_2 = \frac{-27/2 - 0}{81/4 - 9/4} = \frac{-27/2}{72/4} = \frac{-27/2}{18} = -\frac{3}{4}$.Since $m_1 	imes m_2 = (\frac{4}{3}) 	imes (-\frac{3}{4}) = -1$,the angle subtended at the focus is $90^{\circ}$.

The angle subtended by the normal chord at the point $(9, 9)$ on the parabola $y^2 = 9x$ at the focus of the parabola is (in $^{\circ}$)

Similar Questions

The equation of the tangent to the parabola $y^2 = 9x$ which passes through the point $(4, 10)$ is:

$A$ point $P(x, y)$ moves such that its distance from the point $(a, 0)$ is always equal to its distance from the line $x + a = 0$. The locus of the point is:

If $y = mx + c$ is the normal at a point on the parabola $y^2 = 8x$ whose focal distance is $8 \text{ units}$, then $|c|$ is equal to (in $\sqrt{3}$)

Find the point of intersection of the parabola $y^2 = 2px$ and a circle whose center is at the focus of the parabola and which touches the directrix of the parabola.

Find the equation of the normal to the parabola $y^2 = 8x$ at the point $(2, 4)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module