The equations of two sides of a variable triangle are x-0 and y=3, and its third side is a tangent to the parabola y2=6 x. The locus of its circumcentre is:

Question

The equations of two sides of a variable triangle are x-0 and y=3, and its third side is a tangent to the parabola y2=6 x. The locus of its circumcentre is: (A) 4 y2-18 y+3 x+18=0 (B) 4 y2-18 y-3 x+18=0 (C) 4 y2+18 y+3 x+18=0 (D) 4 y2-18 y-3 x-18=0

Tardigrade Admin · Accepted Answer

y2=6 x y2=4 a x ⇒ 4 a=6 ⇒ a=(3/2) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/064cd4ec30e62c33f573749aa1b3a998-.png" /> y = mx +(3/2 m ) ;( m ≠ 0) h =(6 m -3/4 m 2), k =(6 m +3/4 m ), Now eliminating m and we get ⇒ 3 h =2(-2 k 2+9 k -9) ⇒ 4 y 2-18 y +3 x +18=0

Q. The equations of two sides of a variable triangle are $x - 0$ and $y = 3$ , and its third side is a tangent to the parabola $y^{2} = 6 x$ . The locus of its circumcentre is:

$4 y^{2} - 18 y + 3 x + 18 = 0$

$4 y^{2} - 18 y - 3 x + 18 = 0$

$4 y^{2} + 18 y + 3 x + 18 = 0$

$4 y^{2} - 18 y - 3 x - 18 = 0$

$y^{2} = 6 x & y^{2} = 4 a x$
$\Rightarrow 4 a = 6 \Rightarrow a = \frac{3}{2}$

$y = m x + \frac{3}{2 m}; (m \neq = 0)$
$h = \frac{6 m - 3}{4 m ^{2}}, k = \frac{6 m + 3}{4 m}$ , Now eliminating $m$ and we get
$\Rightarrow 3 h = 2 (- 2 k^{2} + 9 k - 9)$
$\Rightarrow 4 y^{2} - 18 y + 3 x + 18 = 0$

Q. The equations of two sides of a variable triangle are x−0 and y=3, and its third side is a tangent to the parabola y2=6x. The locus of its circumcentre is:

4y2−18y+3x+18=0

4y2−18y−3x+18=0

4y2+18y+3x+18=0

4y2−18y−3x−18=0