IGDTUW 2025 — Mathematics PYQ

A straight line of the form $y=mx+c$ is a tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ if $c^2=a^2{m}^2+b^2$. We can transform the given normal form of the line into the slope-intercept form to apply this condition.

Step 1: Rewrite the line in slope-intercept form ($y=mx+c$)

The given line is:

$$x\cos \alpha +y\sin \alpha =p$$

Rearranging to solve for $y$:

$$y\sin \alpha =-x\cos \alpha +p$$

$$y=-\left(\frac{\cos \alpha }{\sin \alpha }\right)x+\frac{p}{\sin \alpha }$$

$$y=(-\cot \alpha )x+p\csc \alpha$$

Comparing this with $y=mx+c$, we have:

• $m=-\cot \alpha$

• $c=p\csc \alpha$

Step 2: Apply the condition of tangency

For an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the condition of tangency is $c^2=a^2{m}^2+b^2$. Substituting our values:

$$(p\csc \alpha {)}^2=a^2(-\cot \alpha {)}^2+b^2$$

$$p^2{\csc }^2\alpha =a^2{\cot }^2\alpha +b^2$$

Step 3: Simplify using trigonometric identities

Recall that ${\csc }^2\alpha =\frac{1}{{\sin }^2\alpha }$ and ${\cot }^2\alpha =\frac{{\cos }^2\alpha }{{\sin }^2\alpha }$:

$$p^2\left(\frac{1}{{\sin }^2\alpha }\right)=a^2\left(\frac{{\cos }^2\alpha }{{\sin }^2\alpha }\right)+b^2$$

Multiply the entire equation by ${\sin }^2\alpha$:

$$p^2=a^2{\cos }^2\alpha +b^2{\sin }^2\alpha$$

Conclusion:

Comparing this result with the given options, the correct choice is (c).