A normal is drawn to the parabola y2 = 9x at the point P(4, 6), S being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

Question

A normal is drawn to the parabola y2 = 9x at the point P(4, 6), S being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is (A) 4 (B) (15/4) (C) 6 (D) (17/4)

Tardigrade Admin · Accepted Answer

Required intercept will be equal to the perpendicular distance from the focus on the tangent at

P

.
Tangent at

P

,

y \cdot 6 = 2 \cdot \frac{9}{4} (x + 4)

\Rightarrow 12 y = 9 x + 36

\Rightarrow 9 x - 12 y + 36 = 0

P = ∣ ∣ \frac{\frac{81}{4} + 36}{81 + 144} ∣ ∣ = ∣ ∣ \frac{225}{4.15} ∣ ∣ = \frac{15}{4}

Q. A normal is drawn to the parabola $y^{2} = 9 x$ at the point $P (4, 6), S$ being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

4

$\frac{15}{4}$

6

$\frac{17}{4}$

Q. A normal is drawn to the parabola y2=9x at the point P(4,6),S being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

4

415​

6

417​

Required intercept will be equal to the perpendicular distance from the focus on the tangent at P. Tangent at P, y⋅6=2⋅49​(x+4) ⇒12y=9x+36 ⇒9x−12y+36=0 P=∣∣​81+144​481​+36​∣∣​=∣∣​4.15225​∣∣​=415​

Q. A normal is drawn to the parabola $y^{2} = 9 x$ at the point $P (4, 6), S$ being the focus, a circle is described on the focal distance of the point P as diameter. The length of the intercept made by the circle on the normal at P is

$\frac{15}{4}$

$\frac{17}{4}$