JEE Corner — Conic Sections for JEE Main and Advanced

The General Circle and Tangency

The chapter so far gives you the standard forms; JEE tests four extensions. This section builds each from what you already know.

1. The general equation of a circle. Expanding $(x-h)^2 + (y-k)^2 = r^2$ always yields the shape

$$x^2 + y^2 + 2gx + 2fy + c = 0$$

Key Point (Reading the general form): centre $(-g, -f)$, radius $\sqrt{g^2 + f^2 - c}$ — a real circle needs $g^2 + f^2 - c > 0$ (equality gives a point circle). Note the equation has equal $x^2, y^2$ coefficients and no $xy$ term.

For $x^2 + y^2 + 6x - 4y + 4 = 0$: $g = 3$, $f = -2$, centre $(-3, 2)$, radius $\sqrt{9 + 4 - 4} = 3$.

2. Position of a point. Write $S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c$. Then $S_1 < 0$, $= 0$, $> 0$ according as the point is inside, on, or outside the circle — the same verdict the squared-distance comparison reaches.

3. Tangency for the circle: distance = radius. The line $y = mx + c$ (or any line) touches a circle exactly when the perpendicular distance from the centre equals the radius — the Chapter 9 distance formula does all the work. For $x^2 + y^2 = r^2$ this packages into $c^2 = r^2(1 + m^2)$.

4. Tangency for the other conics. Substitute $y = mx + c$ into the conic and demand a repeated root (discriminant zero):

Key Point (Tangency conditions): to $y^2 = 4ax$: $c = \frac{a}{m}$. To $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$: $c^2 = a^2m^2 + b^2$. To $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$: $c^2 = a^2m^2 - b^2$.

So every slope $m \ne 0$ gives exactly one tangent to a parabola, $y = mx + \frac{a}{m}$, and every slope gives two parallel tangents to an ellipse, $y = mx \pm \sqrt{a^2m^2 + b^2}$. For the ellipse, the point of contact of $y = mx + c$ is $\left(-\frac{a^2m}{c}, \frac{b^2}{c}\right)$.

Focal Distances, Asymptotes, and Eccentricity Identities

5. Focal distance of a parabola. For $P(x_1, y_1)$ on $y^2 = 4ax$, the focal distance equals the directrix distance:

$$PF = x_1 + a$$

No square roots needed — this one line solves most "distance from the focus" numericals.

6. Focal distances of an ellipse. The derivation of the standard equation produced $PF_1 = a + ex_1$ and $PF_2 = a - ex_1$, summing to $2a$ automatically. On $\frac{x^2}{25} + \frac{y^2}{9} = 1$ at $x_1 = 3$: distances $5 \pm \frac{12}{5} = \frac{37}{5}, \frac{13}{5}$.

7. Asymptotes of a hyperbola. For large $|x|$ the branches of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ hug the lines

$$y = \pm\frac{b}{a}x$$

(the diagonals of the $2a \times 2b$ box). They never meet the curve, and the rectangular (equilateral) hyperbola $a = b$ has perpendicular asymptotes and $e = \sqrt{2}$.

8. Eccentricity identities worth caching.

Ellipse: $b^2 = a^2(1 - e^2)$. Hyperbola: $b^2 = a^2(e^2 - 1)$. A hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and its conjugate $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$ have eccentricities tied by

$$\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$$

since $e_1^2 = \frac{a^2 + b^2}{a^2}$ and $e_2^2 = \frac{a^2 + b^2}{b^2}$.

[JEE Tip] Most conics errors in the exam are sign errors on the three relations: circle needs no relation, ellipse $c^2 = a^2 - b^2$, hyperbola $c^2 = a^2 + b^2$, and the tangency conditions inherit the same sign pattern ($+b^2$ for ellipse, $-b^2$ for hyperbola). Learn them as one table, not four facts.

JEE-Style Solved Examples

Example 1: Reading the general form

Find the centre and radius of $x^2 + y^2 + 6x - 4y + 4 = 0$, and the position of the origin relative to it.

Solution:

Step 1 — Match the general form. $2g = 6$, $2f = -4$: $g = 3$, $f = -2$.

Step 2 — Centre and radius. Centre $(-g, -f) = (-3, 2)$; radius $\sqrt{g^2 + f^2 - c} = \sqrt{9 + 4 - 4} = 3$.

Step 3 — Position of the origin. $S_1 = 0 + 0 + 0 - 0 + 4 = 4 > 0$: outside.

Takeaway: Halve the linear coefficients and negate — the general form reads off in seconds.

Example 2: When is it a circle at all?

For which $c$ does $x^2 + y^2 - 6x + 8y + c = 0$ represent a real circle?

Solution:

Step 1 — Radius formula. $g = -3$, $f = 4$: radius$^2 = 9 + 16 - c = 25 - c$.

Step 2 — Demand positivity. $25 - c > 0$: $c < 25$.

Step 3 — Boundary case. $c = 25$ gives radius 0 — a single point $(3, -4)$.

Takeaway: $g^2 + f^2 - c$ is the gatekeeper: positive → circle, zero → point, negative → nothing.

Example 3: Tangency by distance = radius

For what $k$ is $3x + 4y = k$ tangent to $x^2 + y^2 = 25$?

Solution:

Step 1 — Distance from the centre. $\frac{|3(0) + 4(0) - k|}{\sqrt{9+16}} = \frac{|k|}{5}$.

Step 2 — Set equal to the radius. $\frac{|k|}{5} = 5$: $|k| = 25$.

Step 3 — Both signs. $k = \pm 25$ — one tangent on each side of the circle.

Takeaway: Circle tangency is a Chapter 9 distance computation — no discriminants needed.

Example 4: Tangent to a parabola of given slope

Find the tangent to $y^2 = 8x$ with slope 2.

Solution:

Step 1 — Apply the condition. $a = 2$: tangency needs $c = \frac{a}{m} = \frac{2}{2} = 1$.

Step 2 — Write the tangent. $y = 2x + 1$.

Step 3 — Point of contact. $\left(\frac{a}{m^2}, \frac{2a}{m}\right) = \left(\frac{1}{2}, 2\right)$; check: $2^2 = 8 \cdot \frac{1}{2}$ ✓.

Takeaway: Each slope gives exactly ONE tangent to a parabola — $y = mx + \frac{a}{m}$ is worth memorising whole.

Example 5: Tangents to an ellipse of given slope

Find the tangents to $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with slope 1.

Solution:

Step 1 — Apply the condition. $c^2 = a^2m^2 + b^2 = 16(1) + 9 = 25$.

Step 2 — Both signs. $c = \pm 5$: tangents $y = x + 5$ and $y = x - 5$.

Takeaway: An ellipse admits TWO parallel tangents per slope — they sandwich the curve.

Example 6: Point of contact

Where does $y = x + 5$ touch $\frac{x^2}{16} + \frac{y^2}{9} = 1$?

Solution:

Step 1 — Apply the contact formula. $\left(-\frac{a^2m}{c}, \frac{b^2}{c}\right) = \left(-\frac{16}{5}, \frac{9}{5}\right)$.

Step 2 — Verify on the ellipse. $\frac{256/25}{16} + \frac{81/25}{9} = \frac{16}{25} + \frac{9}{25} = 1$ ✓.

Step 3 — Verify on the line. $-\frac{16}{5} + 5 = \frac{9}{5}$ ✓.

Takeaway: The contact point satisfies BOTH equations — two ten-second checks certify the formula.

Example 7: Focal distance of a parabola point

Find the focal distance of the point on $y^2 = 12x$ with abscissa 4.

Solution:

Step 1 — Identify $a$. $4a = 12$: $a = 3$.

Step 2 — Apply $PF = x_1 + a$. $4 + 3 = 7$.

Takeaway: Focal distance = directrix distance = $x_1 + a$ — the definition doing the arithmetic for you.

Example 8: Focal distances of an ellipse point

Find both focal distances of the point on $\frac{x^2}{25} + \frac{y^2}{9} = 1$ with $x_1 = 3$, and verify the sum.

Solution:

Step 1 — Eccentricity. $c = 4$: $e = \frac{4}{5}$.

Step 2 — Apply $a \pm ex_1$. $5 \pm \frac{4}{5}(3) = 5 \pm \frac{12}{5}$: distances $\frac{37}{5}$ and $\frac{13}{5}$.

Step 3 — Verify. $\frac{37}{5} + \frac{13}{5} = 10 = 2a$ ✓.

Takeaway: $PF = a \pm ex_1$ turns ellipse focal distances into linear arithmetic — and their sum self-checks.

Example 9: Asymptotes and the angle between them

Find the asymptotes of $\frac{x^2}{9} - \frac{y^2}{16} = 1$ and the tangent of the acute angle between them.

Solution:

Step 1 — Write the asymptotes. $y = \pm\frac{b}{a}x = \pm\frac{4}{3}x$.

Step 2 — Angle between slopes $\pm\frac{4}{3}$. $\tan\theta = \left|\frac{\frac{4}{3} + \frac{4}{3}}{1 - \frac{16}{9}}\right| = \frac{8/3}{7/9}$.

Step 3 — Simplify. $\tan\theta = \frac{8}{3} \cdot \frac{9}{7} = \frac{24}{7}$.

Takeaway: Asymptote slopes are $\pm\frac{b}{a}$ — the Chapter 9 angle formula does the rest.

Example 10: Conjugate hyperbola eccentricities

Show that the eccentricities $e_1$ of $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $e_2$ of its conjugate satisfy $\frac{1}{e_1^2} + \frac{1}{e_2^2} = 1$.

Solution:

Step 1 — Write both eccentricities. $e_1^2 = \frac{a^2 + b^2}{a^2}$; the conjugate swaps the roles: $e_2^2 = \frac{a^2 + b^2}{b^2}$.

Step 2 — Add the reciprocals. $\frac{a^2}{a^2 + b^2} + \frac{b^2}{a^2 + b^2} = 1$. ∎

Takeaway: Both hyperbolas share the same $c^2 = a^2 + b^2$ — the identity is that shared focus circle in disguise.

Example 11: A tangent to a hyperbola

For which $c$ is $y = 2x + c$ tangent to $\frac{x^2}{9} - \frac{y^2}{16} = 1$?

Solution:

Step 1 — Apply the condition. $c^2 = a^2m^2 - b^2 = 9(4) - 16 = 20$.

Step 2 — Both signs. $c = \pm 2\sqrt{5}$: tangents $y = 2x \pm 2\sqrt{5}$.

Step 3 — Slope audit. The condition needs $a^2m^2 > b^2$, i.e. $|m| > \frac{b}{a} = \frac{4}{3}$ — satisfied by $m = 2$ ✓.

Takeaway: Hyperbola tangents of slope $m$ exist only when $|m|$ beats the asymptote slope $\frac{b}{a}$.

Example 12: Circle through three points (general form at work)

Find the circle through $(0, 0)$, $(4, 0)$ and $(0, 6)$.

Solution:

Step 1 — Feed the points to the general form. $(0,0)$: $c = 0$; $(4, 0)$: $16 + 8g = 0$, so $g = -2$; $(0, 6)$: $36 + 12f = 0$, so $f = -3$.

Step 2 — Write. $x^2 + y^2 - 4x - 6y = 0$.

Step 3 — Anatomy. Centre $(2, 3)$, radius $\sqrt{4 + 9} = \sqrt{13}$.

Takeaway: Three points → three linear equations in $g, f, c$ — the general form exists exactly for this job.