NIMCET 2020 — Mathematics PYQ

Step 1: Convert the ellipse equation to standard form

The given equation of the ellipse is:

$$x^2+16y^2=16$$

Divide both sides by $16$ to get the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$:

$$\frac{x^2}{16}+\frac{16y^2}{16}=\frac{16}{16}$$

$$\frac{x^2}{16}+\frac{y^2}{1}=1$$

By comparing this with the standard equation, we find:

• $a^2=16$

• $b^2=1$

Step 2: Find the slope ($m$) of the tangent

The tangent line makes an angle of $60^{\circ }$ with the x-axis. The slope ($m$) of a line is given by $m=\tan (\theta )$:

$$m=\tan (60^{\circ })=\sqrt{3}$$

Step 3: Apply the condition of tangency

The equation of a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in slope form is:

$$y=mx\pm \sqrt{a^2{m}^2+b^2}$$

Substitute the values of $a^2$, $b^2$, and $m$ into the formula:

$$y=\sqrt{3}x\pm \sqrt{16(\sqrt{3}{)}^2+1}$$

$$y=\sqrt{3}x\pm \sqrt{16(3)+1}$$

$$y=\sqrt{3}x\pm \sqrt{48+1}$$

$$y=\sqrt{3}x\pm \sqrt{49}$$

$$y=\sqrt{3}x\pm 7$$

This gives us two possible equations for the tangent:

1. $y=\sqrt{3}x+7⟹\sqrt{3}x-y+7=0$

2. $y=\sqrt{3}x-7⟹\sqrt{3}x-y-7=0$

Conclusion

Comparing our findings with the given choices, the correct equation present in the options is:

$$\sqrt{3}x-y+7=0$$

Correct Answer: C