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Conics · eight-question check

Read the idea, work independently, then explain what changed.

TOPIC 01

Conics

Try two tasks per skill independently, then revisit the linked skill.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Foundation#Your turn

Choose the major semiaxis length for x²/9+y²/25=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $3$
Option B $5$
Option C $9$
Option D $25$

Working and explanation

BUILD THE REASONING

Hint 1

The larger denominator identifies the major axis.

Hint 2

Take its square root.

Worked solution

Compare 9 and 25.
$a^2=25$
Convert squared length to length.
$a=5$

B: 5.

Checks and common pitfalls: A full major axis would have length 10.

Think first. Reveal a hint when the class is ready.

02 / Foundation#Your turn

Choose c² for x²/25+y²/9=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $34$
Option B $16$
Option C $4$
Option D $25$

Working and explanation

BUILD THE REASONING

Hint 1

This is an ellipse with positive squared terms.

Hint 2

Subtract minor from major squared semiaxis.

Worked solution

Apply the focal relation.
$c^2=a^2-b^2$
Calculate the difference.
$25-9=16$

B: 16.

Checks and common pitfalls: The focus lies inside the major vertex.

Think first. Reveal a hint when the class is ready.

03 / Foundation#Your turn

Choose c² for x²/9−y²/16=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $7$
Option B $25$
Option C $5$
Option D $-7$

Working and explanation

BUILD THE REASONING

Hint 1

A focus lies beyond a vertex.

Hint 2

Use a²+b².

Worked solution

Apply the hyperbola relation.
$c^2=9+16$
Add the parameters.
$c^2=25$

B: 25.

Checks and common pitfalls: The ellipse subtraction formula does not apply.

Think first. Reveal a hint when the class is ready.

04 / Foundation#Your turn

Choose the positive-slope asymptote of x²/9−y²/16=1.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $y=3x/4$
Option B $y=4x/3$
Option C $y=16x/9$
Option D $y=4x/3+5$

Working and explanation

BUILD THE REASONING

Hint 1

Remove the constant term.

Hint 2

Take the positive square-root branch.

Worked solution

Set the homogeneous quadratic to zero.
$y^2/16=x^2/9$
Use the required positive slope.
$y=4x/3$

Checks and common pitfalls: The asymptote slopes depend on semiaxes, not their squares.

Think first. Reveal a hint when the class is ready.

05 / Foundation#Your turn

Choose the focus of y²=8x.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $(8,0)$
Option B $(4,0)$
Option C $(2,0)$
Option D $(0,2)$

Working and explanation

BUILD THE REASONING

Hint 1

Compare with y²=4ax.

Hint 2

The axis is horizontal.

Worked solution

Read a=8/4.
$a=2$
Put a on the x-axis.
$F=(2,0)$

Checks and common pitfalls: State which standard-form convention is being used.

Think first. Reveal a hint when the class is ready.

06 / Foundation#Your turn

Choose the vertex of (y−2)²=4(x+3).

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $(3,2)$
Option B $(-3,2)$
Option C $(-2,3)$
Option D $(-3,-2)$

Working and explanation

BUILD THE REASONING

Hint 1

Set both shifted coordinates to zero.

Hint 2

Solve x+3=0 and y−2=0.

Worked solution

The square is zero at y=2.
$y=2$
The right side is zero at x=−3.
$x=-3$

B: (−3,2).

Checks and common pitfalls: A vertex is not generally the focus.

Think first. Reveal a hint when the class is ready.

07 / Foundation#Your turn

Choose all intersections of y²=4x and y=x.

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $(4,4)$
Option B $(0,0)$
Option C $(0,0),(4,4)$
Option D $(0,0),(4,-4)$

Working and explanation

BUILD THE REASONING

Hint 1

Factor x(x−4)=0.

Hint 2

Every point must satisfy both equations.

Worked solution

Retain both roots.
$x=0,4$
Use y=x for each.
$(x,y)=(0,0),(4,4)$

Checks and common pitfalls: An intersection is a solution of a system, not of just the eliminated equation.

Think first. Reveal a hint when the class is ready.

08 / Foundation#Your turn

After line substitution the equation is 0x²+2x−4=0. How many real intersections follow?

Skills and prerequisite lessons

Scope: JM01 / JM02

Option A $0$
Option B $1$
Option C $2$
Option D $\infty$

Working and explanation

BUILD THE REASONING

Hint 1

Identify the actual degree first.

Hint 2

Solve 2x−4=0.

Worked solution

The equation is linear.
$x=2$
A line gives one corresponding ordinate.

B: one intersection; tangency does not follow from this alone.

Checks and common pitfalls: The discriminant criterion requires a nonzero quadratic coefficient.

Think first. Reveal a hint when the class is ready.