SATHEE: Parabola 3 Question 2

Parabola 3 Question 2

2. If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is

(a) 3

$(2000, 2 M)$

(b) 9

(d) -3

Match the Columns

Show Answer

Answer:

Correct Answer: 2. (b)

Solution:

Given, equation of parabola is $x^{2} = 4 y$

and the chord is $x - \sqrt{2} y + 4 \sqrt{2} = 0$

From Eqs. (i) and (ii), we have

$\begin{aligned} [\sqrt{2} (y - 4)]^{2} & = 4 y \\ \Rightarrow & 2 (y - 4)^{2} & = 4 y \\ \Rightarrow & (y - 4)^{2} & = 2 y \\ \Rightarrow & y^{2} - 8 y + 16 & = 2 y \\ \Rightarrow & y^{2} - 10 y + 16 & = 0 \end{aligned}$

Let the roots of Eq. (iii) be $y_{1}$ and $y_{2}$ Then,

$y_{1} + y_{2} = 10 and y_{1} y_{2} = 16$

Again from Eqs. (i) and (ii), we have

$\begin{aligned} x^{2} & = 4 \frac{x}{\sqrt{2}} + 4 \\ \Rightarrow x^{2} - 2 \sqrt{2} x - 16 & = 0 \end{aligned}$

Let the roots of Eq. (v) be $x_{1}$ and $x_{2}$

Then, $x_{1} + x_{2} = 2 \sqrt{2}$

and $x_{1} x_{2} = - 16$

Clearly, length of the chord $A B$

$\begin{aligned} = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}} \\ = \sqrt{{(x_{1} + x_{2})}^{2} - 4 x_{1} x_{2} + {(y_{1} + y_{2})}^{2} - 4 y_{1} y_{2}} \\ = \sqrt{8 + 64 + 100 - 64} [∵ (a - b)^{2} = (a + b)^{2} - 4 a b] \\ = \sqrt{108} = 6 \sqrt{3} [from Eqs. (iv) and (vi)] \end{aligned}$

Parabola 3 Question 2

2. If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Parabola 3 Question 2

2. If x+y=k is normal to y2=12x, then k is

Match the Columns

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu

2. If $x + y = k$ is normal to $y^{2} = 12 x$ , then $k$ is