Mathematics · Differential Equations

JEE Main 2024 — 6 April, Shift 1 — Question 22

Let a conic CC pass through the point (4,−2)(4,-2) and P(x,y),x≥3P(x, y), x \geq 3, be any point on CC. Let the slope of the line touching the conic C only at a single point P be half the slope of the line joining the points P and (3,−5)(3,-5). If the focal distance of the point (7,1)(7,1) on C is d , then 12 d equals \qquad

Answer: 75

Numerical answer — enter this value.

Step-by-step solution

P(x,y)&x≥3\quad \mathrm{P}(\mathrm{x}, \mathrm{y}) \& \mathrm{x} \geq 3

Slope of line at P(x,y)P(x, y) will be dydx=12(y+5x−3)\frac{d y}{d x}=\frac{1}{2}\left(\frac{y+5}{x-3}\right)

⇒2dy(y+5)=1(x−3)dx\Rightarrow 2 \frac{d y}{(y+5)}=\frac{1}{(x-3)} d x

⇒2ln⁡(y+5)=ln⁡(x−3)+C\Rightarrow 2 \ln (\mathrm{y}+5)=\ln (\mathrm{x}-3)+\mathrm{C}

Passes through (4,−2)(4,-2)

⇒2ln⁡(3)=ln⁡(1)+C\Rightarrow 2 \ln (3)=\ln (1)+C

⇒C=2ln⁡(3)\Rightarrow \mathrm{C}=2 \ln (3)

⇒2ln⁡(y+5)=ln⁡(x−3)+2ln⁡(3)\Rightarrow 2 \ln (\mathrm{y}+5)=\ln (\mathrm{x}-3)+2 \ln (3)

⇒2(ln⁡(y+53))=ln⁡(x−3)\Rightarrow 2\left(\ln \left(\frac{\mathrm{y}+5}{3}\right)\right)=\ln (\mathrm{x}-3)

⇒(y+53)2=(x−3)\Rightarrow\left(\frac{\mathrm{y}+5}{3}\right)^{2}=(\mathrm{x}-3)

⇒(y+5)2=9(x−3)\Rightarrow(\mathrm{y}+5)^{2}=9(\mathrm{x}-3)

Parabola 4a=94 \mathrm{a}=9

a=94a=\frac{9}{4}

d=(74)2+62d=\sqrt{\left(\frac{7}{4}\right)^{2}+6^{2}}

d=6254d=\frac{\sqrt{625}}{4}

d=254\mathrm{d}=\frac{25}{4}

12 d=7512 \mathrm{~d}=75

Solution figure

Answer key and solution verified before publishing.

Practise Differential Equations

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Exam
JEE Main 2024
Subject
Mathematics
Chapter
Differential Equations
Topic
Methods of solving a First Order,First Degree Differential
Let a conic C pass through the point (4,-2) and P(x, y), x geq 3 , be… | JEE Main 2024 PYQ with Solution · DhiX AI