Question 29/40: Set 2 - Try Out - Mathematics (TKA)

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If a function $y = \sqrt{x^2 - 7}$ , then....

$y = \frac{4}{3}x - \frac{7}{3}$ is the equation of the tangent line at $x = 4$
The curve is a circle centered at $(0,0)$
The line $y = -\frac{3}{4}x + 6$ intersects perpendicularly with the tangent line at $x = 4$
$y = \frac{4}{3}x - \frac{25}{3}$ is the tangent line to the curve at $(4, -3)$

ONLY $(1), (2), (3)$ are correct

ONLY $(1)$ and $(3)$ are correct

ONLY $(2)$ and $(4)$ are correct

ONLY $(4)$ is correct

ALL statements are correct

Given the function $y = \sqrt{x^2 - 7}$ . The derivative of this function is

y' = \frac{x}{\sqrt{x^2 - 7}}

The ordinate of the point on the curve with abscissa $4$ is

f(4) = \sqrt{4^2 - 7} = \sqrt{9} = 3

The gradient of the tangent line at point $(4,3)$ can be found by substituting $x = 4$ into the derivative.

y' = f'(x) = \frac{x}{\sqrt{x^2 - 7}}

f'(4) = \frac{4}{\sqrt{4^2 - 7}} = \frac{4}{\sqrt{9}} = \frac{4}{3}

With gradient $m = \frac{4}{3}$ and point $(4,3)$ , the equation of the tangent line is

y - y_1 = m(x - x_1)

y - 3 = \frac{4}{3}(x - 4)

y - 3 = \frac{4}{3}x - \frac{16}{3}

y = \frac{4}{3}x - \frac{7}{3}

The first statement is correct.

For values of $x$ and $y$ that satisfy the condition, square both sides.

y = \sqrt{x^2 - 7}

y^2 = \left(\sqrt{x^2 - 7}\right)^2

y^2 = x^2 - 7

x^2 - y^2 = 7

The form $x^2 - y^2 = 7$ is not the equation of a circle.

The second statement is incorrect.

The line $y = -\frac{3}{4}x + 6$ has gradient $m_1 = -\frac{3}{4}$ .

From the first statement, the gradient of the tangent line at $x = 4$ is $m_2 = \frac{4}{3}$ .

Check if both lines are perpendicular by multiplying both gradients.

m_1 \cdot m_2 = -\frac{3}{4} \cdot \frac{4}{3} = -1

Since $m_1 \cdot m_2 = -1$ , both lines are perpendicular.

The third statement is correct.

Check if the curve passes through point $(4, -3)$ by substituting the value $x = 4$ .

y = \sqrt{4^2 - 7} = \sqrt{9} = 3

The curve does not pass through point $(4, -3)$ , but passes through point $(4,3)$ .

The fourth statement is incorrect.

Therefore, the correct statements are the first and third.

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