Question

Q. Choose the correct answer.
Q. Which of the following statement is true about the curve defined by 6y2=x(x-2)
i). The curve passes through (0,0) AND X=0 IS TANGENT LINE AT (0,0)
ii) The curve is symmetric about x-axis
iii) The curve is symmetric about y-axis
iv) The curve lies to the left of y-axis
a. i) and iii)
b. i) and ii)
c. . ii) and iii)
d. iii) and iv)

Answer 1

Answer to Question #91574 – Math – Geometry

Question

Which of the following statement is true about the curve defined by $6y^2 = x(x-2)$:

i). The curve passes through (0,0) AND $x=0$ IS TANGENT LINE AT (0,0)

ii) The curve is symmetric about $x$-axis

iii) The curve is symmetric about $y$-axis

iv) The curve lies to the left of $y$-axis

a. i) and iii)

b. i) and ii)

c. . ii) and iii)

d. iii) and iv)

Solution

Given equation of curve

$$6y^2 = x(x-2)$$

Now considering option i)

On substituting (0,0)

$$6 \times 0^2 = 0 \times (x-2)$$ $$0 = 0$$

(0,0) satisfies this equation therefore this curve passes through (0,0)

We know that

Slope of tangent $\frac{dy}{dx}$

On differentiating equation of curve

$$6 \times 2y \times \frac{dy}{dx} = 2x-2$$ $$12y \times \frac{dy}{dx} = 2(x-1)$$ $$\frac{dy}{dx} = \frac{2(x-1)}{12y}$$ $$\frac{dy}{dx} = \frac{x-1}{6y}$$

At (0,0) slope of tangent $\frac{dy}{dx}$ at $(0,0) = \frac{-1}{0} = -\infty$

Therefore at origin this curve has vertical tangent at $x=0$. Therefore $x=0$ is tangent to the curve at origin.

Thus, the option i) is correct.

Now considering option ii)

We can see that

On replacing $y$ by $-y$ equation of curve becomes

$$6(-y)^2 = x(x-2)$$ $$6y^2 = x(x-2)$$

Since we can see that equation of curve doesn't change on replacing $y$ by $-y$ therefore we can say that curve is symmetric about $x$-axis.

Thus, option ii) is correct.

Now considering option iii)

We can see that

On replacing $x$ by $-x$ equation of curve becomes

$$6(y)^2 = -x(-x-2)$$ $$6y^2 = x(x+2)$$

Since we can see that equation of curve does change on replacing $x$ by $-x$ therefore we can say that curve is not symmetric about $x$-axis.

Thus, the option iii) is incorrect.

Now considering option iv)

We can see that

For $x > 2$

Curve will be defined as $(x-2) > 0$ and $x > 0$

Therefore

$x(x-2) > 0$ so $y$ will have real values for all $x > 0$

therefore curve lies to the right of $y$-axis also.

Thus, the option iv) is incorrect.

OPTION B) i) and ii) IS CORRECT.

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