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QJEEM24Ma69ca: JEE Main 2024 Mathematics - Class 11 JEEM24Ma69ca Solution: Let be the lines passing through the point and touching the…

Let L1,L2L_1, L_2 be the lines passing through the point P(0,1)P(0,1) and touching the parabola 9x2+12x+18y−14=09 x^2+12 x+18 y-14=0 . Let QQ and RR be the points on the lines L1L_1 and L2L_2 such that the △PQR\triangle P Q R is an isosceles triangle with base QRQ R . If the slopes of the lines QRQ R are m1m_1 and m2m_2 , then 16(m12+m22)16\left(m_1^2+m_2^2\right) is equal to __________.

Step-by-Step Verified Solution

amp;9x2+12x+18y−14=0amp;(x+23)2=−2(y−1)…(1)\begin{aligned}& 9 x^2+12 x+18 y-14=0 \\& \left(x+\frac{2}{3}\right)^2=-2(y-1) \ldots(1)\end{aligned}

Equation of tangent to (1)

amp;t(x+23)=−(y−1)+12t2 passes through (0,1)amp;⇒23t=12t2⇒t=0,43⇒m=0,m=−43\begin{aligned}& t\left(x+\frac{2}{3}\right)=-(y-1)+\frac{1}{2} t^2 \text { passes through }(0,1) \\& \Rightarrow \frac{2}{3} t=\frac{1}{2} t^2 \Rightarrow t=0, \frac{4}{3} \Rightarrow m=0, m=-\frac{4}{3}\end{aligned}

JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Parabola Question 11 English Explanation

amp;⇒∣m+431−43m∣=∣m∣⇒∣3m+43−4m∣=∣m∣amp;⇒3m+4=m(3−4m) or 3m+4=−m(3−4m)amp;3m+4=3m−4m2 or 3m+4=−3m+4m2amp;4m2+4=0 (not possible) or 4m2−6m−4=0amp;m1+m2=32,m1m2=−1amp;⇒m12+m22=174amp;16(m12+m22)=68\begin{aligned}& \Rightarrow\left|\frac{m+\frac{4}{3}}{1-\frac{4}{3} m}\right|=|m| \Rightarrow\left|\frac{3 m+4}{3-4 m}\right|=|m| \\& \Rightarrow 3 m+4=m(3-4 m) \text { or } 3 m+4=-m(3-4 m) \\& 3 m+4=3 m-4 m^2 \text { or } 3 m+4=-3 m+4 m^2 \\& 4 m^2+4=0 \text { (not possible) or } 4 m^2-6 m-4=0 \\& m_1+m_2=\frac{3}{2}, m_1 m_2=-1 \\& \Rightarrow m_1^2+m_2^2=\frac{17}{4} \\& 16\left(m_1^2+m_2^2\right)=68\end{aligned}

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