Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Let $\mathrm{A}$ be the point $(1,2)$ and $\mathrm{B}$ be any point onthe curve $\mathrm{x}^2+y^2=16$ . If the centre of the locus of the point $\mathrm{P},$ which divides the line segment $\mathrm{AB}$ in the ratio $3: 2$ is the point $\mathrm{C}(\alpha, \beta)$ then the length of the line segment $\mathrm{AC}$ is
Option: 1
$\frac{6 \sqrt{5}}{5}$

Option: 2
$\frac{2 \sqrt{5}}{5}$

Option: 3
$\frac{3 \sqrt{5}}{5}$

Option: 4
$\frac{4 \sqrt{5}}{5}$

Answers (1)

best_answer

$\frac{12 \cos \theta+2}{5}=h \Rightarrow 12 \cos \theta=5 \mathrm{~h}-2$

sq & add

$\begin{aligned} & 144=(5 h-2)^2+(5 k-4)^2 \\ & \left(x-\frac{2}{5}\right)^2+\left(y-\frac{4}{5}\right)^2=\frac{144}{25} \\ & \text { Centre } \equiv\left(\frac{2}{5}, \frac{4}{5}\right) \equiv(\alpha, \beta) \\ & A C=\sqrt{\left(1-\frac{2}{5}\right)^2+\left(2-\frac{4}{5}\right)^2} \\ & =\sqrt{\frac{9}{25}+\frac{36}{25}}=\frac{\sqrt{45}}{5}=\frac{3 \sqrt{5}}{5} \end{aligned}$

Posted by

shivangi.bhatnagar

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl. If pK_b of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75

Trending Articles/News

anam-khan

JEE Main 2025: BTech ECE cut-offs for female candidates go up at top NITs

anam-khan

JEE Main 2025 Session 2 City Slip (OUT) Live: Admit card likely on March 29 at jeemain.nta.nic.in; cut-offs

anam-khan

AESL expands Aakash Digital 2.0 to offer AI-powered coaching for NEET, JEE, Olympiads

Latest Question