```html

Cone Slice View

https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/Xsw59cREc3RhrqIumCmFIAp5WfhmKaiN5foVhnT2.png

Tilted plane slicing a cone produces an ellipse.

Ellipse from a Tilted Plane

A right circular cone cut by a slanted plane, within a single nappe, creates an ellipse.

More tilt makes the ellipse slimmer; less tilt makes it closer to a circle.

Key Points:

  • Plane angle is steeper than the base but flatter than the cone side.
  • Cut stays within one nappe, giving a closed, oval curve.
  • Changing the tilt alters size and eccentricity, not the conic type.
``````html

Focus Magic

https://asset.sparkl.ac/pb/sparkl-vector-images/img_ncert/HwfCYJEMztL3zlpxxbxZMTJvwvqTz3vpdB6QfXxv.png

Ellipse with foci \(F_1\) and \(F_2\)

Constant-Sum Locus

An ellipse is the locus of all points \(P\) such that \(PF_1 + PF_2\) is constant.

That unchanging sum defines the curve’s size—move \(P\) anywhere and the value stays fixed.

Key Points:

  • Foci \(F_1\) and \(F_2\) stay fixed inside the ellipse.
  • Constant sum \(PF_1 + PF_2 = k\).
  • Quiz: If one measurement gives \(10\text{ cm}\), what is \(k\) for all points?
```

Key Distances

\[c^{2}=a^{2}-b^{2}\]

Variable Definitions

\(a\) Semi-major axis length
\(b\) Semi-minor axis length
\(c\) Focal length (centre to focus)

Applications

Find Foci

Given \(a\) and \(b\), compute \(c\) to locate the two foci.

Check Validity

Test if lengths satisfy \(c^{2}=a^{2}-b^{2}\) to confirm an ellipse exists.

Equation Build-Up

1
\[\sqrt{(x+c)^2+y^{2}}+\sqrt{(x-c)^2+y^{2}}=2a\]

Distance-sum rule for an ellipse with foci \((\pm c,0)\); major axis length \(2a\).

2
\[2x^{2}+2y^{2}+2c^{2}+2\sqrt{\bigl[(x+c)^2+y^{2}\bigr]\bigl[(x-c)^2+y^{2}\bigr]}=4a^{2}\]

Square once to begin eliminating radicals; collect like terms.

3
\[b^{2}x^{2}+a^{2}y^{2}=a^{2}b^{2}\]

Isolate the remaining radical, square again, and substitute \(b^{2}=a^{2}-c^{2}\).

4
\[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\]

Divide by \(a^{2}b^{2}\) to reach the standard equation centred at the origin.

Key Insight:

The link \(b^{2}=a^{2}-c^{2}\) ties focal spacing to axis lengths, sealing the derivation.

```html

Ellipse Plot

Graph of x²⁄9 + y²⁄4 = 1

What the graph shows

This graphical view plots the ellipse \( \frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 \) on the coordinate plane.

The curve crosses the x-axis at \((\pm3,0)\) and the y-axis at \((0,\pm2)\).

These intercepts equal the semi-major axis \(a = 3\) and semi-minor axis \(b = 2\) introduced earlier.

Key Points:

  • Graphical view makes the oval shape clear.
  • Ends on x-axis: \((\pm3,0)\).
  • Ends on y-axis: \((0,\pm2)\).
```

Test Yourself

Question

For an ellipse with semi-major axis \(a\) and focal distance \(c\), which formula gives the semi-minor axis \(b\)?

1
\(b^{2}=a^{2}+c^{2}\)
2
\(b^{2}=a^{2}-c^{2}\)
3
\(b^{2}=c^{2}-a^{2}\)
4
\(b^{2}=2ac\)

Hint:

Think of a right-angle triangle formed by \(a, b,\) and \(c\) at the centre of the ellipse.

Ellipse Wrap-Up

Quick recap: an ellipse is the set of points whose summed distances to two foci is constant.

Standard form: \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \).

Major axis length \(2a\); minor axis \(2b\); focal distance satisfies \(c^2 = a^2 - b^2\).

Eccentricity \( e=\frac{c}{a} \) gauges stretch; for every ellipse \( 0 < e < 1 \).

The larger denominator marks the major axis, revealing horizontal or vertical orientation.

Translate or rotate the form to describe any ellipse you encounter in practice.

Thank You!

We hope you found this lesson informative and engaging.