if the point $(2, 3)$ lies on $k x^{2} - 3 y^{2} + 2 x + y - 2 = 0$ then $k$ is equal to

$\frac{1}{17}$

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$16$

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$7$

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$12$

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Solution

The correct option is C
$7$

Explantion for the correct options:
Point on curve:
Substitute $(2, - 3)$ in the given equation.
Given $(2, 3)$ lies on $k x^{2} - 3 y^{2} + 2 x + y - 2 = 0$
$k \times 2^{2} - 3 \times 3^{2} + 2 \times 2 - 3 - 2 = 0$
$\Rightarrow 4 k - 27 + 4 - 5 = 0$
$\Rightarrow 4 k = 28$
$\Rightarrow k = 7$
Hence option $(C)$ is the correct answer.

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Similar questions

State, true or false :

(i) the line $\frac{x}{2} + \frac{y}{3} = 0$ passes through the point (2, 3).

(ii) the line $\frac{x}{2} + \frac{y}{3} = 0$ passes through the point (4, -6).

(iii) the point (8, 7) lies on the line y - 7 = 0

(iv) the point (-3, 0) lies on the line x + 3 = 0

(v) if the point (2, a) lies on the line 2x - y = 3, then a = 5.

k x^{2} - 3 y^{2} + 2 x + y - 2 = 0

P

Q (1, - 2, 3)

\frac{x}{1} = \frac{y}{4} = \frac{z}{5}

P

2 x + 3 y - 4 z + 22 = 0

P Q

If the point (2,a) lies on the line 2x-y=3 then 'a' is equal to

(2, 3)

(4, 5)

y - 4 x + 3 = 0

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