JEE Main 2023 — Mathematics Coordinate Geometry

Given,

$a{x}^{2}+2bx+cy=0$ax2+2bx+cy=0 and ${d}^{2}+2ex+fy=0$d2+2ex+fy=0 intersect on the line $y=1$y=1,

And $a,b,c$a,b,c are in G.P.

So, ${b}^{2}=ac$b2=ac

Now putting the value of $b$b in $a{x}^{2}+2bx+cy=0$ax2+2bx+cy=0and taking $y=1$y=1 we get,

$$a{x}^{2}+2bx+c=0$$ax2+2bx+c=0

$$\Rightarrow {\mathrm{ax}}^{2}+2\sqrt{\mathrm{ac}x}+c=0(\because {b}^{2}=\mathrm{ac})$$⇒ax2+2acx​+c=0(∵b2=ac)

$$\Rightarrow (x\sqrt{a}+\sqrt{c}{)}^{2}=0$$⇒(xa​+c​)2=0

$\Rightarrow x=-\frac{\sqrt{c}}{\sqrt{a}}$⇒x=−a​c​​ and ${x}^{2}=\frac{c}{a}.......(1)$x2=ac​.......(1)

Now, putting the value of x&{x}^{2} in $d{x}^{2}+2ex+fy=0$dx2+2ex+fy=0 and taking $y=1$y=1 we get,

$$\Rightarrow d(\frac{c}{a})+2e[-\frac{\sqrt{c}}{\sqrt{a}}]+f=0$$⇒d(ac​)+2e[−a​c​​]+f=0

$$\Rightarrow \frac{dc}{a}+f=2e\sqrt{\frac{c}{a}}$$⇒adc​+f=2eac​​

$$\Rightarrow \frac{d}{a}+\frac{f}{c}=2e\sqrt{\frac{1}{ac}}$$⇒ad​+cf​=2eac1​​

$\Rightarrow \frac{d}{a}+\frac{f}{c}=\frac{2e}{b}$⇒ad​+cf​=b2e​ as $[b=\sqrt{ac}]$[b=ac​]

$\therefore \frac{d}{a},\frac{e}{b},\frac{f}{c}$∴ad​,be​,cf​are in $A.P.$A.P.