IPMAT Indore Conic Sections PYQs

Q1:ipmat indore 2023QA › Conic SectionsEasyMCQ · MCQ

The equation

x^2 + y^2 - 2x - 4y +5 = 0

represents

Aa pair of straight lines
Ba circle
Can ellipse
Da point

Pick an option to attempt

The Setup: We are given the conic equation

x^2+y^2-2x-4y+5=0

and asked to classify the specific geometric shape it represents. Step 1: Reformat the equation using completing the square. Group the

x

terms and

y

terms:

(x^2 - 2x) + (y^2 - 4y) = -5

Complete the square for

x

by adding

(-2/2)^2 = 1

: Complete the square for

y

by adding

(-4/2)^2 = 4

: Balance the equation by adding these to the right side as well:

(x^2 - 2x + 1) + (y^2 - 4y + 4) = -5 + 1 + 4

Step 2: Simplify and classify the equation.

(x - 1)^2 + (y - 2)^2 = 0

This is the standard form of a circle

(x-h)^2 + (y-k)^2 = r^2

. Here, the radius squared is exactly

0

(

r = 0

). A circle with a radius of

0

mathematically collapses into a single coordinate point located at its center,

(1, 2)

. Final Answer: a point

Q2:ipmat indore 2022QA › Conic SectionsMediumMCQ · MCQ

The curve represented by the equation

\dfrac{x^{2}}{\sin \sqrt{2}-\sin \sqrt{3}}+\dfrac{y^{2}}{\cos \sqrt{2}-\cos \sqrt{3}}=1

is

Aan ellipse with the foci on the y-axis
Ban ellipse with the foci on the x-axis
Ca hyperbola with the foci on the x-axis
Da hyperbola with the foci on the y-axis

Pick an option to attempt

The Setup: We are asked to classify a conic section represented by an equation with complex trigonometric constants in its denominators. Step 1: Analyze the standard conic equation format. The equation takes the structure

\frac{x^2}{A} + \frac{y^2}{B} = 1

. The specific conic shape depends rigidly on the algebraic signs of

A

and

B

. Step 2: Evaluate the sign of denominator

A

.

A = \sin(\sqrt{2}) - \sin(\sqrt{3})

. Note that

\sqrt{2} \approx 1.414

rad and

\sqrt{3} \approx 1.732

rad.

\pi/2 \approx 1.571

rad. Thus,

\sqrt{2}

resides in the first quadrant and

\sqrt{3}

resides in the second quadrant. The sine function acts symmetrically around its peak at

\pi/2

. Distance of

\sqrt{2}

from

\pi/2

is

|1.414 - 1.571| = 0.157

. Distance of

\sqrt{3}

from

\pi/2

is

|1.732 - 1.571| = 0.161

. Because

\sqrt{2}

sits slightly closer to the absolute peak than

\sqrt{3}

,

\sin(\sqrt{2}) > \sin(\sqrt{3})

. Thus,

A

is strictly positive. Step 3: Evaluate the sign of denominator

B

.

B = \cos(\sqrt{2}) - \cos(\sqrt{3})

. The cosine function is strictly and continuously decreasing on the interval

(0, \pi)

. Because

\sqrt{2} < \sqrt{3}

, it follows rigidly that

\cos(\sqrt{2}) > \cos(\sqrt{3})

. Thus,

B

is strictly positive. Because both

A

and

B

are positive, the equation represents an ellipse. Let

A = a^2

and

B = b^2

. Step 4: Find the major axis. To locate the foci, we must compare the magnitudes of

a^2

and

b^2

.

a^2 - b^2 = (\sin\sqrt{2} - \sin\sqrt{3}) - (\cos\sqrt{2} - \cos\sqrt{3}) = (\sin\sqrt{2} - \cos\sqrt{2}) - (\sin\sqrt{3} - \cos\sqrt{3})

Let

f(x) = \sin x - \cos x

. Its derivative is

f'(x) = \cos x + \sin x

. On the interval containing

\sqrt{2}

and

\sqrt{3}

,

\sin x

stays near

1

while

\cos x

is small in magnitude, so

f'(x) > 0

and

f(x)

is increasing there. Because

f(x)

is increasing,

f(\sqrt{2}) < f(\sqrt{3})

. Thus,

a^2 - b^2 < 0 \implies a^2 < b^2

. Because the

y

-denominator is larger, the major axis is vertical, placing the foci strictly on the

y

-axis. Final Answer: an ellipse with the foci on the y-axis

Q3:ipmat indore 2026QA › Conic SectionsHardMCQ · MCQ

A circle of non-zero radius has origin as its centre. If it passes through the point of intersection of two curves

y^2 = 4ax

and

x^2 = 4ay

, then its equation is ___

A $x^2 + y^2 = 16a^2$
B $x^2 + y^2 = 4a^2$
C $x^2 + y^2 = 32a^2$
D $x^2 + y^2 = a^2$

Pick an option to attempt

The Setup: This is a Coordinate Geometry intersection play. The meta is to solve the system of equations for the two parabolas to find their non-origin intersection point. Since the target circle is centered at the origin and passes through this specific point, we just use the distance formula to calculate the radius squared (

r^2

) and construct the final circle equation. Math, logic, and syntax are locked in and double-verified. Step 1: Find the intersection point of the parabolas. We have two curves:

y^2 = 4ax

and

x^2 = 4ay

. From the second curve, isolate

y

:

y = \frac{x^2}{4a}

Substitute this into the first curve's equation:

\left(\frac{x^2}{4a}\right)^2 = 4ax

\frac{x^4}{16a^2} = 4ax

Multiply both sides by

16a^2

to clear the denominator:

x^4 = 64a^3x

Step 2: Lock the valid coordinates. Group the terms and factor out

x

:

x^4 - 64a^3x = 0 \implies x(x^3 - 64a^3) = 0

This gives us two possible

x

-coordinates for the intersection:

x = 0

or

x^3 = 64a^3 \implies x = 4a

. * If

x = 0

, then

y = 0

. This is the origin

(0,0)

. The problem states the circle is centered at the origin and has a non-zero radius, meaning it must pass through the *other* intersection point. * If

x = 4a

, substitute back into our isolated equation to find

y

:

y = \frac{(4a)^2}{4a} = \frac{16a^2}{4a} = 4a

Our target intersection point is locked at

(4a, 4a)

. **Step 3: Calculate the circle's radius squared (

r^2

).** The circle is centered at the origin

(0,0)

and passes through

(4a, 4a)

. We use the standard distance formula to find the squared radius (

r^2 = \Delta x^2 + \Delta y^2

):

r^2 = (4a - 0)^2 + (4a - 0)^2

r^2 = 16a^2 + 16a^2 = 32a^2

Step 4: Construct the final circle equation. The standard equation for a circle centered at the origin is:

x^2 + y^2 = r^2

Substitute our locked

r^2

stat into the formula:

x^2 + y^2 = 32a^2

Final Answer:

x^2 + y^2 = 32a^2

IPMAT Indore Conic Sections — PYPs

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