If ax^2+2hxy+by^2-82x+98y+144=0 is the equati | vedclass

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(D) For a parabola,the distance from any point $P(x, y)$ on the parabola to the focus $S$ is equal to the perpendicular distance from $P$ to the direc

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two coincident lines

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(D) For a parabola,the distance from any point $P(x, y)$ on the parabola to the focus $S$ is equal to the perpendicular distance from $P$ to the directrix $L$.$PS = PN$$\Rightarrow \sqrt{(x-2)^2+(y+3)^2} = \left|\frac{3x-2y+5}{\sqrt{3^2+(-2)^2}}\right|$$\Rightarrow (x-2)^2+(y+3)^2 = \frac{(3x-2y+5)^2}{13}$$\Rightarrow 13(x^2-4x+4+y^2+6y+9) = 9x^2+4y^2+25-12xy-20y+30x$$\Rightarrow 13x^2-52x+52+13y^2+78y+117 = 9x^2+4y^2-12xy+30x-20y+25$$\Rightarrow 4x^2+12xy+9y^2-82x+98y+144 = 0$Comparing this with the given equation $ax^2+2hxy+by^2-82x+98y+144=0$,we get $a=4, h=6, b=9$.Now,the equation $ax^2+2hxy+by^2=0$ becomes $4x^2+12xy+9y^2=0$.This can be written as $(2x+3y)^2=0$.Since the discriminant $h^2-ab = 6^2-(4)(9) = 36-36=0$,it represents two coincident lines.

If $ax^2+2hxy+by^2-82x+98y+144=0$ is the equation of a parabola with focus $(2,-3)$ and directrix $3x-2y+5=0$,then $ax^2+2hxy+by^2=0$ represents

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